Let's Denest A Difficult Radical | Viewer Suggested

Here is a derivation of a denesting formula for radicals of the form shown in the video. It relies on the idea that the square root inside the radical can form a basis orthogonal to any other value that does not include the square root as a factor. To derive the formula, start by assuming for rational f and g : {1} cbrt(a + d*sqrt(s)) = cbrt(f) + cbrt(g)*sqrt(s) Raise this equation to power 3 : a + d*sqrt(s) = f + g*s*sqrt(s) + 3*cbrt(f*g)*sqrt(s)*(cbrt(f) + cbrt(g)*sqrt(s)) a + d*sqrt(s) = f + 3*s*cbrt(f*g^2) + sqrt(s)*(s*g + 3*cbrt(g*f^2)) Equate the coefficients of sqrt(s) and the remainder that have no sqrt(s) to get a = f + 3*s*cbrt(f*g^2) d = s*g + 3*cbrt(g*f^2) = cbrt(g) * (s*cbrt(g^2) + 3*cbrt(f^2)) Eliminate the terms on the left by adding (-a/d)*d 0 = f + 3*s*cbrt(f*g^2) + (-a/d)*( s*g + 3*cbrt(g*f^2) ) Multiply by d/g 0 = d * f/g + 3*s*d*cbrt(f/g) - a*s - 3*a*cbrt(f^2 /g^2) Let x = cbrt(f/g) which means cbrt(g) = cbrt(f) / x {2} 0 = d*x^3 - 3*a*x^2 + 3*s*d*x - a*s Return to the equations for a and d, solve the a equation for cbrt(g^2) cbrt(g^2) = (a - f) / (3*s*cbrt(f)) Substitute cbrt(g^2) into the equation for d d = cbrt(g) * (s * (a - f) / (3*s*cbrt(f)) + 3*cbrt(f^2)) d = 1/3 *cbrt(g/f) *(a + 8*f) Substitute x = cbrt(f/g) d = 1/3 *(a + 8*f) / x Solve for f f = 1/8 *(3*x*d - a) Use this result and cbrt(g) = cbrt(f) / x , to write formula {1} as cbrt(a + d*sqrt(s)) = cbrt(f) * (1 + sqrt(s) / x ) {3} cbrt(a + d*sqrt(s)) = 1/2 * cbrt(3*x*d - a) * (1 + sqrt(s) / x) where x is a rational root of {2} , since if x is a root that is not rational, the equality still holds but it will not be denested.

Nice problem and solution. Other approach. Let x=cbrt(70-22sqrt(7)). Multiply both sides by cbrt(7) and compare to a-b×sqrt(7): cbrt(7)x = cbrt(490-154sqrt(7)) = a-b×sqrt(7) The conjugate: cbrt(490+154sqrt(7)) = a+b×sqrt(7) After multiplying we get: a^2-7b^2 = cbrt(490^2-154^2×7)=cbrt(7^3×700-7^3×484)=7cbrt(216)=7×6=42 So a^2-7b^2=42, hence a=7 and b=1. We get cbrt(7)x = 7-sqrt(7), hence x = (7-sqrt(7))/cbrt(7) = cbrt(49)-7^(1/6).

Nice and clever solution.

Very nice problem and thanks to viewer @XJWill1 for contributing this. If you know the original source of this problem, then please let me know. I tried this without first watching the video of course and this had me stumped for a while. It was immediately clear that ³√(70 − 22√7) cannot be expressed as x − √y for any rational x and y, as @XJWill1 pointed out. But before I discuss my approach I would like to point out the following. Let a be a nonzero rational number, b a positive rational number with √b irrational and suppose there exists a rational number x as well as a positive rational number y with √y irrational such that (1) ³√(a + √b) = x + √y then we must also have (2) ³√(a − √b) = x − √y Multiplying (1) and (2) we have (3) ³√(a² − b) = x² − y Now, the right hand side of (3) is rational, so the left hand side of (3) must also be rational which implies that a² − b is indeed the cube of a rational number. This means that a² − b being the cube of a rational number is a necessary condition for the denestability of ³√(a ± √b) with rational a and b as x ± √y with rational x and y. But the converse is not true which means that a² − b being the cube of a rational number is not sufficient to guarantee denestability of ³√(a ± √b) as x ± √y with rational x and y. Example: for ³√(3 + √8) we have a = 3, b = 8, so a² − b = 9 − 8 = 1 = 1³ is the cube of a rational number, but ³√(3 + √8) is not denestable as x + √y with rational x and y. With this out of the way, here is how I approached this problem. Let me make it clear up front that I do not recommend my initial solution since the approach shown by SyberMath in the video is definitely the way to go. But I think it is worth sharing for additional insight into this problem, so here it goes. The nested cube root ³√(70 − 22√7) is reminiscent of the type of expressions we encounter in the formulas for the roots of a depressed cubic with one real root and two conjugate complex roots. Specifically, if we have a depressed cubic x³ + px + q = 0 where p and q are real numbers and D = (¹⁄₂q)² + (¹⁄₃p)³ is positive then this cubic equation has one real and two conjugate complex roots which are given by x₁ = u + v x₂ = −¹⁄₂·(u + v) + ¹⁄₂√3·(u − v)·i x₃ = −¹⁄₂·(u + v) − ¹⁄₂√3·(u − v)·i where u = ³√(−¹⁄₂q + √D) v = ³√(−¹⁄₂q − √D) Since we can rewrite 22√7 as √(22²·7) = √3388 my approach was to start by constructing a depressed cubic equation which has ³√(70 + 22√7) + ³√(70 − 22√7) = ³√(70 + √3388) + ³√(70 − √3388) as its sole real root. This is easily done, since we must evidently have −¹⁄₂q = 70 and therefore q = −140, and D = (¹⁄₂q)² + (¹⁄₃p)³ = 3388 then gives (¹⁄₃p)³ = 3388 − 4900 = −1512 = −2³·3³·7, so p = −18·³√7. Therefore, the depressed cubic equation x³ − 18·³√7·x − 140 = 0 has x₁ = ³√(70 + 22√7) + ³√(70 − 22√7) as its only real root. Obviously we are not interested per se in finding an alternative expression for the sum of u = ³√(70 + 22√7) and v = ³√(70 − 22√7) but we do want to find an alternative expression for v = ³√(70 − 22√7) itself. So, how do we do that? Well, if we would know all three roots of our cubic equation, then we would know both u + v and u − v and then we could easily find v (and u of course). But solving the cubic equation in the usual way will not give us what we want because this will only express the roots in terms of ³√(70 + 22√7) and ³√(70 − 22√7). What we need is an alternative way to find the real root x₁ = u + v of our cubic equation, because we can then factor out (x − x₁) which will give us a quadratic equation which we can solve to get the two conjugate complex roots x₂ and x₃ which will then give us u − v. The difficulty with x³ − 18·³√7·x − 140 = 0 of course is the irrational coefficient −18·³√7 of the linear term. If we could devise a substitution to get rid of this and get a cubic equation with rational coefficients then we could use the rational root theorem to find out if the new equation has a rational root. Since we can multiply 18·³√7 by ³√(7²) to get 18·³√(7³) = 18·7 = 126 and since (³√(7²))³ = 7² = 49 is also rational, an obvious substitution is x = ³√49·z which turns x³ − 18·³√7·x − 140 = 0 into 49z³ − 126z − 140 = 0 and dividing both sides by 7 this gives 7z³ − 18z − 20 = 0 It is not hard to see that z = 2 is a solution of this equation, and factoring out (z − 2) we have (z − 2)(7z² + 14z + 10) = 0 z − 2 = 0 ⋁ 7z² + 14z + 10 = 0 z − 2 = 0 ⋁ z² + 2z + ¹⁰⁄₇ = 0 z − 2 = 0 ⋁ (z + 1)² = −³⁄₇ z = 2 ⋁ z = −1 + √(³⁄₇)·i ⋁ z = −1 − √(³⁄₇)·i Since x = ³√49·z and since ³√49·√(³⁄₇) = ³√(7²)·√(³⁄₇) = ⁶√(7⁴)·⁶√(3³/7³) = ⁶√(3³·7) = ⁶√(3³)·⁶√7 = √3·⁶√7 the roots of the cubic equation x³ − 18·³√7·x − 140 = 0 are therefore x₁ = 2·³√49 x₂ = −³√49 + √3·⁶√7·i x₃ = −³√49 − √3·⁶√7·i and comparing this with x₁ = u + v x₂ = −¹⁄₂·(u + v) + ¹⁄₂√3·(u − v)·i x₃ = −¹⁄₂·(u + v) − ¹⁄₂√3·(u − v)·i where u = ³√(70 + 22√7) v = ³√(70 − 22√7) we can see that we have u + v = 2·³√49 u − v = 2·⁶√7 which gives u = ³√49 + ⁶√7 v = ³√49 − ⁶√7 so we have ³√(70 + 22√7) = ³√49 + ⁶√7 ³√(70 − 22√7) = ³√49 − ⁶√7 Note that for the cubic equation 7z³ − 18z − 20 = 0 or z³ − ¹⁸⁄₇z − ²⁰⁄₇ = 0 we have p = −¹⁸⁄₇, q = −²⁰⁄₇ which gives D = (¹⁄₂q)² + (¹⁄₃p)³ = (−¹⁰⁄₇)² + (−⁶⁄₇)³ = ¹⁰⁰⁄₄₉ − ²¹⁶⁄₃₄₃ = ⁷⁰⁰⁄₃₄₃ − ²¹⁶⁄₃₄₃ = ⁴⁸⁴⁄₃₄₃ so √D = ²²⁄₄₉√7 which gives the roots z₁ = u + v z₂ = −¹⁄₂·(u + v) + ¹⁄₂√3·(u − v)·i z₃ = −¹⁄₂·(u + v) − ¹⁄₂√3·(u − v)·i where u = ³√(¹⁰⁄₇ + ²²⁄₄₉√7), v = ³√(¹⁰⁄₇ − ²²⁄₄₉√7). But since we have already found that z₁ = 2 z₂ = −1 + √(³⁄₇)·i z₃ = −1 − √(³⁄₇)·i this means that u + v = 2, u − v = ²⁄₇√7 and therefore u = 1 + ¹⁄₇√7, v = 1 − ¹⁄₇√7 so we have ³√(¹⁰⁄₇ + ²²⁄₄₉√7) = 1 + ¹⁄₇√7 ³√(¹⁰⁄₇ − ²²⁄₄₉√7) = 1 − ¹⁄₇√7 Since x = ³√49·z it follows that ³√(70 + 22√7) = ³√49·³√(¹⁰⁄₇ + ²²⁄₄₉√7) = ³√49·(1 + ¹⁄₇√7) = ³√49 + ⁶√7 ³√(70 − 22√7) = ³√49·³√(¹⁰⁄₇ − ²²⁄₄₉√7) = ³√49·(1 − ¹⁄₇√7) = ³√49 − ⁶√7 so it is indeed possible to denest ³√(70 − 22√7) by taking out a factor ³√49 and then denest ³√(¹⁰⁄₇ − ²²⁄₄₉√7) = 1 − ¹⁄₇√7 in the conventional way. Alternatively, we could take out a factor ³√(¹⁄₇) which gives ³√(70 + 22√7) = ³√(¹⁄₇)·³√(490 + 154√7) = ³√(¹⁄₇)·(7 + √7) = ³√49 + ⁶√7 ³√(70 − 22√7) = ³√(¹⁄₇)·³√(490 − 154√7) = ³√(¹⁄₇)·(7 − √7) = ³√49 − ⁶√7 so it is also possible to denest ³√(70 − 22√7) by taking out a factor ³√(¹⁄₇) and then denest ³√(490 − 154√7) = 7 − √7 in the conventional way. This corresponds with a substitution x = ³√(¹⁄₇)·y which turns the equation x³ − 18·³√7·x − 140 = 0 into y³ − 126y − 980 = 0 which has the real solution y₁ = ³√(490 + 154√7) + ³√(490 − 154√7) = (7 + √7) + (7 − √7) = 14. The solution presented in the video corresponds with a substitution x = ⁶√7·w which turns the equation x³ − 18·³√7·x − 140 = 0 into w³ − 18w − 20√7 = 0 which has the real solution w₁ = ³√(10√7 + 22) + ³√(10√7 − 22) = (√7 + 1) + (√7 − 1) = 2√7 but obviously this substitution is less convenient as it does not result in a cubic with rational coefficients and a rational solution.

the usual approach works. you consider that the solution is of the form a-bsqrt(7), cube and get the set of 2 equations in a and b. you change variable to u=a/b and with a little effort you get a single 3rd degree equation in u. the only real solution to that equation is u=7. you resubstitute in the 2 equations and find a and b.

As it turns out, there actually is a formula for denesting ³√(a ± b√c) with rational a and b and positive rational c, √c irrational, which is ³√(a ± b√c) = ¹⁄₂·³√(3bt − a)·(1 ± (1/t)·√c) where t is a rational root of bt³ − 3at² + 3bct − ac = 0 The idea behind this is clear, taking out a factor such that the value of the remaining nested cubic root has a rational part ¹⁄₂, which makes sense because that means taking out a factor k from the radicand such that ³√(a + b√c) + ³√(a − b√c) = ³√k. This formula works both for denesting denestable cube roots of quadratic binomial surds which can be denested using only square roots and for denesting denestable cube roots of quadratic binomial surds where this is not possible. For example, to denest ³√(90 − 34√7) we have a = 90, b = 34, c = 7 and we need to find a rational root of 34t³ − 270t² + 714t − 630 = 0 which is t = 3 and this gives ³√(3bt − a) = ³√216 = 6 so we have ³√(90 − 34√7) = ¹⁄₂·6·(1 − ¹⁄₃·√7) = 3 − √7 To denest ³√(70 − 22√7) we have a = 70, b = 22, c = 7 and we need to find a rational root of 22t³ − 210t² + 462t − 490 = 0 which is t = 7 and this gives ³√(3bt − a) = ³√392 = ³√(8·49) = ³√8·³√49 = 2·³√49 so we have ³√(70 − 22√7) = ¹⁄₂·2·³√49·(1 − ¹⁄₇·√7) = ³√49 − ³√49·√(¹⁄₇) = ³√49 − ⁶√(7⁴)·⁶√(1/7³) = ³√49 − ⁶√7 So, the formula for denesting a denestable nested cube root ³√(a ± b√c) with rational a and b and positive rational c, √c irrational which I found on the net evidently works, but why does it work and how can we derive it? First observe that if we take out a factor k from the radicand a ± b√c where we assume k to be rational then we have (1) ³√(a ± b√c) = ³√(k·(a/k ± (b/k)·√c)) = ³√k·³√(a/k ± (b/k)·√c) If ³√(a/k ± (b/k)·√c) where a/k and b/k are rational can be denested into a binomial quadratic surd then there exist rational numbers x and y such that ³√(a/k ± (b/k)·√c) = x ± y·√c and then we will have ³√(a/k + (b/k)·√c) + ³√(a/k − (b/k)·√c) = 2x Clearly, if 2x is not 1, then we can multiply k by (2x)³ and k will still be rational since x is rational, and then with the new value of k we will have ³√(a/k + (b/k)·√c) + ³√(a/k − (b/k)·√c) = 1 This means that if ³√(a/k ± (b/k)·√c) is denestable into a binomial quadratic surd for some rational value of k, then it will always be possible to select a rational value of k such that (2) ³√(a/k ± (b/k)·√c) = ¹⁄₂·(1 ± (1/t)·√c) for some nonzero rational value of t. Cubing both sides this gives (3) a/k ± (b/k)·√c = ¹⁄₈·(1 + 3c/t²) ± ¹⁄₈·(3/t + c/t³)·√c where a, b, c, k, t are rational and √c is irrational. In accordance with the theorem that states that if the sum or difference of a rational quantity and an irrational quantity is equal to the sum or difference of another rational quantity and an irrational quantity then both the rational quantities and the irrational quantities must be equal this implies that we have (4a) a/k = ¹⁄₈·(1 + 3c/t²) (4b) b/k = ¹⁄₈·(3/t + c/t³) Dividing (4a) by (4b) we have (5) a/b = (1 + 3c/t²)/(3/t + c/t³) and cross multiplying (5) and then multiplying both sides by t³ we obtain (6) bt³ − 3at² + 3bct − ac = 0 which means that in order for ³√(a ± b√c) with rational a, b, c, √c irrational to be denestable, there must exist a rational value of t which satisfies this cubic equation. Multiplying both sides of (4a) by k and multiplying both sides of (4b) by 3kt we have (7a) a = ¹⁄₈·k·(1 + 3c/t²) (7b) 3bt = ¹⁄₈·k·(9 + 3c/t²) and subtracting (7a) from (7b) this gives (8) 3bt − a = k Multiplying both sides of (2) by ³√k we have ³√k·³√(a/k ± (b/k)·√c) = ¹⁄₂·³√k·(1 ± (1/t)·√c) and in accordance with (1) and (8) this gives (9) ³√(a ± b√c) = ¹⁄₂·³√(3bt − a)·(1 ± (1/t)·√c) which completes the proof of the formula.

Rather strange 7 and 22! Shades of pi.

Suppose - 22 + 10√7 = (u + √v)³ where u & v are rational & v is not the square of a rational. (u² - v)³ = (u + √v)³.(u - √v)³ = (- 22 + 10√7).(- 22 - 10√7) = 484 - 700 = - 216 = (-6)³, so u and v exists. If we did not get the cube of a rational number, then u and v would not have existed. Now - 22 + 10√7 = (u³ + 3uv) + (3u² + v)√v. So - 22 = (u³ + 3uv) and 10√7 = (3u² + v)√v. ∴ (u-√v)³ = (u³+3uv) - (3u²+v)√v = - 22 - 10√7. So - 216 = 484 - 700 = (- 22 +10√7).(- 22 -10√7) = [(u + √v)³].[(u - √v)³] = (u² - v)³. Hence (u² - v) = ∛(- 216)) = - 6 & so v = u² - (-6) = u² + 6. Therefore - 22 = (u³ + 3u.v) = u³ + 3u.(u² + 6) = 4.u³ + 18u. So 4.u³ + 18.u + 22 = 0. By the Rational Root theorem, u = -1 because 4.u³ + 18.u + 22 = 2(u - 1).(2u² - 2u + 11) and the quadratic factor has no real roots. Thus v = u² + 6 = 7. So ∛(-22 + 10√7) = (u+√v) = -1 + √7. ∴ ∛{70√7 - 22} = ∛{(√7).(-22 + 10√7)} = 7¹ᐟ⁶. (-1 + √7) = 7¹ᐟ⁶.(7¹ᐟ² - 1) = 7⁴ᐟ⁶ - 7¹ᐟ⁶ = 7²ᐟ³ - 7¹ᐟ⁶. .

Very cunning.