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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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    Try $t = x - 3.$ – 2017-01-02
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    $x=2, x=4$ are solutions. – 2017-01-02
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    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. – 2017-01-02

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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    why don't just substitute $t = x - 4$ instead of doing $t = x - 3$ first and $y = t - 1$ after that? – 2017-01-02
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    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. – 2017-01-02
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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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    "Id est" doesn't mean "therefore". – 2017-01-02
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    @Rob Arthan "Id est" it's "i.e." – 2017-01-02
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    Quite so. "i.e." isn't a good way to proclaim your answer. – 2017-01-02
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    @Rob Arthan I don't agree with you. – 2017-01-02
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    OK. Let's agree to differ. – 2017-01-02
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    @RobArthan `i.e.` literally means `that is`. It's generally accepted as an inference in constructs like `x^2=0 i.e. x=0`. – 2017-01-02
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    @dxiv: I know exactly what the Latin means and I think it is poor style to use it in the way you suggest: something like "$\log_e\,x = 1$. i.e., $\ln\,x = 1$", would be OK: "i.e." is a reminder that $\log_e$ and $\ln$ are the same thing. Using "i.e" to signify an inference is misleading. – 2017-01-02