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Solve $(x-2)^6+(x-4)^6=64$ using a substitution.

I tired using $t=x-2$.

But again I have to know the expansion of the power $6$.

Can it be transformed to a quadratic equation ?

I know that it can be somehow solved by expanding the powers. But I'm trying to get a good transformation by a sub.

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    $\begingroup$ Try $t = x - 3.$ $\endgroup$ Jan 2, 2017 at 0:34
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    $\begingroup$ $x=2, x=4$ are solutions. $\endgroup$
    – Crostul
    Jan 2, 2017 at 0:34
  • $\begingroup$ Why do you want to solve it this way ? Because as I see it, I don't think you can make it quadratic with any substitution, the best you can do is an equation of degree 3. $\endgroup$
    – Blencer
    Jan 2, 2017 at 0:55

3 Answers 3

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If you put $t = x-3$, you have $$ (t-1)^6 + (t+1)^6 = 64 $$ the odd terms cancel out when you expand the LHS, so you get $$ u^3 + 15u^2 + 15u + 1 = 32 $$

where $u = (x-3)^2$. From looking at either the initial equation or this one, it's clear that $u = 1$ (corresponding to $t=\pm1$ and $x = 4,2$) is a solution, so can factor to $$ (u^2 + 16u + 31)(u-1) = 0 $$

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Put $y=x-4$

$$(y+2)^6+y^6=2^6$$

It should be clear that $0$ and $-2$ are solutions.

The function $f(y)=(y+2)^6+y^6-2^6$ is decreasing for $y \in (-\infty,-1)$. And increasing for $(-1,\infty)$, so it at most has two roots.

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    $\begingroup$ why don't just substitute $t = x - 4$ instead of doing $t = x - 3$ first and $y = t - 1$ after that? $\endgroup$ Jan 2, 2017 at 1:07
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    $\begingroup$ How does the substitution help? We could say exactly the same things of the original equation, and I'm not sure the substitution really clarifies things. $\endgroup$ Jan 2, 2017 at 1:14
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Let $x-2=a$ and $4-x=b$.

Hence, $a+b=2$ and $a^6+b^6=64$ or $$(a^2+b^2)(a^4+b^4-a^2b^2)=64$$ or $$(2-ab)(a^2b^2-16ab+16)=32$$ or $$ab(a^2b^2-18ab+48)=0.$$

$ab=0$ gives $x=2$ or $x=4$. $ab=9-\sqrt{33}$ does not give a real roots and $ab=9+\sqrt{33}$ does not give a real roots.

Id est, the answer is $\{2,4\}$.

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  • $\begingroup$ "Id est" doesn't mean "therefore". $\endgroup$
    – Rob Arthan
    Jan 2, 2017 at 1:16
  • $\begingroup$ @Rob Arthan "Id est" it's "i.e." $\endgroup$ Jan 2, 2017 at 1:18
  • $\begingroup$ Quite so. "i.e." isn't a good way to proclaim your answer. $\endgroup$
    – Rob Arthan
    Jan 2, 2017 at 1:23
  • $\begingroup$ @Rob Arthan I don't agree with you. $\endgroup$ Jan 2, 2017 at 1:25
  • $\begingroup$ OK. Let's agree to differ. $\endgroup$
    – Rob Arthan
    Jan 2, 2017 at 1:32

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