Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.

Question

Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.

Answer:

$a+b = 7, ab = 2$

$$\begin{align} (a+b)^6 &= a^6 + 6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6 \\[4pt] a^6 + b^6 &= (a+b)^6 - (6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5) \\ &= (a+b)^6 - (6ab(a^4 + b^4) + 15a^2b^2 (a^2 + b^2) + 20(ab)^3) \end{align}$$

now, $$\begin{align} a^4 + b^4 &= (a+b)^4 - (4a^3b + 6a^2b^2 + 4ab^3) \\ &= (a+b)^4 - (4ab(a^2 + b^2) + 6(ab)^2) \\ &= (a+b)^4 - (4ab((a + b)^2 - 2ab) + 6(ab)^2) \\ &= 7^4 - (4(2)(7^2 - 2(2)) + 6(2)^2) \\ &= 2017 \end{align}$$

so

$$\begin{align} &\phantom{=}\; (a+b)^6 - (6ab(a^4 + b^4) + 15a^2b^2 (a^2 + b^2) + 20(ab)^3)\\ &= 7^6 - (6\cdot2\cdot(2017) + 15(2)^2 (7^2 - 2(2)) + 20(2)^3) \\ &= 90585 \end{align}$$ correct?

Answer 1

Why work so hard? From the quadratic formula we have:

$\{ a,b \} = \left\{ \frac{7-\sqrt{41}}{2}, \frac{7+\sqrt{41}}{2} \right\}$ so $a^6 + b^6 = 90585$.

---

My goodness!!! A downvote on a simple, clear, correct answer.

Well, if the approach demands that the solution be done "by hand"—which was NOT part of the problem statement!— then here:

$$a^6 + b^6 = \frac{1}{2^6} \left[ ((7 - \sqrt{41})^2 )^3 + ((7 + \sqrt{41})^2 )^3\right]$$

$$= \frac{1}{64} \left[ (90 - 14 \sqrt{41})^3 + (90 + 14 \sqrt{41})^3\right]$$

Now cancel the terms with odd powers of $(14 \sqrt{41})$ (because they have opposite signs in the pairs), leaving

$$\frac{1}{64} \left[ 2 \cdot 90^3 + 6 \cdot 90 \cdot 14^2 \cdot 41\right] = 90585$$

Satisfied?

Answer 2

Holy unsimplified long expressions Batman. $$ \begin{align} a^2+b^2 &=(a+b)^2-2ab\\ &=45 \end{align} $$ and $$ \begin{align} a^6+b^6 &= (a^2+b^2)^3-3(ab)^2(a^2+b^2)\\ &= (a^2+b^2)((a^2+b^2)^2-3(ab)^2)\\ &= 45\times(45^2-12)\\ &=90585\\ \end{align} $$

Answer 3

A better way to handle this is to observe the factorization $$a^6 + b^6 = (a^2 + b^2)(a^4 - a^2 b^2 + b^4),$$ from which we observe $$a^2 + b^2 = (a+b)^2 - 2ab,$$ and $$a^4 - a^2 b^2 + b^4 = (a^2 + b^2)^2 - 3a^2 b^2.$$

Since we know that $$x^2 - 7x + 2 = (x - a)(x - b) = x^2 - (a+b)x + ab,$$ it follows that $a+b = 7$ and $ab = 2$. So we just put everything together:

$$a^2 + b^2 = 7^2 - 2(2) = 49 - 4 = 45,$$ $$a^4 - a^2 b^2 + b^4 = (45)^2 - 3(2)^2 = 2013,$$

therefore $$a^6 + b^6 = 45(2013) = 90585.$$

Answer 4

For either root of the equation:

• $x^2 = 7x - 2$

• $x^4 = (7x - 2)^2 = 49 x^2 - 28 x + 4 = 49(7x-2) - 28 x + 4 = 315 x - 94$

• $x^6 = (7x - 2)(315 x - 94) = 2205 x^2 - 1288 x + 188 = 14147 x - 4222$

It follows that $\,a^6+b^6 = 14147 (a+b) - 2 \cdot 4222 = 14147 \cdot 7 - 8444 = 90585\,$.

---

[ EDIT ] $\;$ The underlying idea in the above is based on the Euclidean division:

$$ x^6 = (x^4 + 7 x^3 + 47 x^2 + 315 x + 2111) \cdot(x^2 - 7 x + 2) + (14147 x - 4222) $$

Then substituting $x = a,b$ the product is zero, which leaves only the remainder to evaluate, and adding the two equalities gives the same result as in the last line of the answer.

The shortcut here was to derive the remainder directly, without doing the full long division and calculating the quotient, which is not needed in this case.

Answer 5

You have the correct answer, but here's a shorter way.

From Vieta's formulas, $a + b = 7$ and $ab = 2$

$$a^2 + b^2 = (a + b)^2 - 2ab = 7^2 - 2(2) = 49-4 = 45$$ $$a^4 + b^4 = (a^2 + b^2)^2 - 2(ab)^2 =45^2 - 2(2^2) = 2017$$ $$a^6 + b^6 = (a^4 + b^4)(a^2 + b^2) - (ab)^2(a^2 + b^2) = 2017 \times 45 - 2^2(45) = 90585$$

Answer 6

Given $a$ and $b$ be the roots of the equation $x^2-7x+2=0$. Sum of roots, $$a+b=7$$ Product of roots, $$ab=2.$$ Now, $(a-b)^2=(a+b)^2-4ab=49-8=41$.

$\Rightarrow a-b=\pm\sqrt{41}$. Taking $a-b=\sqrt{41}$. On solving equation $a+b=7$ and $a-b=\sqrt{41}$. We obtain, $a=\frac{7+\sqrt{41}}{2}$ and $b=\frac{7-\sqrt{41}}{2}.$

Answer 7

From Vieta’s rules, $a+b=7, ab=2$. $$a^3+b^3=(a+b)(a^2+b^2-ab)=(a+b)((a+b)^2-3ab)= 7(49-2\cdot 3)=7\cdot 43=301 $$ Again, $$(a^3+b^3)^2=a^6+b^6+2a^3b^3$$$$301^2=a^6+b^6+2(2)^3$$ so $$a^6+b^6=301^2-16=90585. $$

Answer 8

Alternatively, let $x_n = a^n + b^n$. Then $x_{n+2}=7x_{n+1}-2x_n$ and so we get

$x_0=2$

$x_1=7$

$x_2=7x_1-2x_0=45$

$x_3=7x_2-2x_1=301$

$x_4=7x_3-2x_2=2017$

$x_5=7x_4-2x_3=13517$

$x_6=7x_5-2x_4=90585$

This is easy to do by hand.

Answer 9

What the hey, I'll toss in one more (there are plenty of ways to work with the binomials, and some of the other answers already given are ones I'd have tried first):

We need $$ \ a + b \ = \ 7 \ \ , \ \ a - b \ = \ \sqrt{\Delta} \ = \ \sqrt{41} \ \ , \ \ ab \ = \ 2 \ \ , $$ $$ a^2 + b^2 \ = \ (a + b)^2 - 2ab \ = \ 49 - 4 \ = \ 45 \ \ , $$ as already presented. We can write $$ (a + b)^6 \ + \ (a - b)^6 \ \ = \ \ 2a^6 \ + \ 2b^6 \ + \ 30·(a^4b^2 \ + \ a^2b^4) $$ $$ = \ \ 2·(a^6 \ + \ b^6) \ + \ 30·(ab)^2·(a^2 \ + \ b^2) \ \ = \ \ 2·(a^6 \ + \ b^6) \ + \ 30·2^2·45 $$

$$ = \ \ 7^6 \ + \ 41^3 \ \ = \ \ 49^3 \ + \ 41^3 \ \ . $$

If we need to do this "by hand", we have $ \ 49^3 + 41^3 \ = \ (49 + 41)·(49^2 - 49·41 + 41^2) \ \ , $ for which we can use little "binomial tricks" to produce $ \ (50 - 1)^2 \ - \ (50 - 1)·(40 + 1) \ + \ (40 + 1)^2 $ $ = \ 2401 - 2009 + 1681 \ = \ 2073 \ \Rightarrow \ 49^3 + 41^3 \ = \ 90·2073 \ = \ 207,300 - 20,730 $ $ = \ 186,570 \ \ . $ We also have $ \ 30·4·45 \ = \ 6·4·15^2 \ = \ 6·(4·225) \ = \ 6·900 \ = \ 5400 \ \ . $

Thus, $$ 2·(a^6 \ + \ b^6) \ \ = \ \ [ \ (a + b)^6 \ - \ (a - b)^6 \ ] \ - \ 30·(ab)^2·(a^2 \ + \ b^2) $$ $$ = \ \ 186,570 \ - \ 5400 \ \ = \ \ 181,170 \ \ \Rightarrow \ \ a^6 \ + \ b^6 \ \ = \ \ 90,585 \ \ . $$

[This is somewhere between what you started out doing and Dan's method.]