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I was solving some problems and I came across this question:

Q: The number of real roots of $(x+3)^4 + (x+5)^4 = 16$ is

                            (a) 0            (b) 2
                            (c) 4            (d) none of these

Solution: put $y = x + (3+5)/2 = x+4$

    the equation becomes

=> $(y-1)^4 + (y+1)^4 = 16$ ---- (i)
=> $2{y^4 + 6(y)^2 + 1 } = 16$ --------(ii)

My question is how was (i) converted to (ii)? I just couldn't get it. Please help?

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Are $x$ and $X$ different or the same? – Ross Millikan Feb 21 '13 at 6:03
I have edited your question on the assumption that $x$ and $X$ both should be $x$ (my edits may take a little while to show). If not you should clarify. As for getting from (i) to (ii), simply expand the fourth powers out and add the results term by term (basically as one of the answers already suggests). – Glen_b Feb 21 '13 at 6:05

4 Answers 4

$$(y-1)^4+(y+1)^4$$ $$=y^4-\binom 41y^3+\binom 42y^2-\binom 43y+1+(y^4+\binom 41y^3+\binom 42y^2+\binom 43y+1)$$ $$=2(y^4+6y^2+1)$$

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no no my question is how was equation (i) converted to equation (ii) i am stuck there further i know how to do it please help me in that part – Dimensionless Feb 21 '13 at 6:02
@Akash, do you know binomial theorem? – lab bhattacharjee Feb 21 '13 at 6:05
thank you very much – Dimensionless Feb 21 '13 at 6:12

$$ \begin{align} (y−1)^4 +(y+1)^4 &= \left(y^4 − \binom{4}{1}y^3 + \binom{4}{2}y^2 − \binom{4}{3}y + 1\right) \\ &+ \left(y^4 + \binom{4}{1}y^3 + \binom{4}{2}y^2 + \binom{4}{3}y + 1\right) \\ &=2(y^4 + 6y^2 + 1) \end{align} $$

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Hi, welcome to the site. Our website supports MathJax for outputting better looking and more legible mathematics. Please try to edit your question using the MathJax capabilities. As it stands it is a bit hard to tell what you mean to write. – Willie Wong Feb 21 '13 at 8:01
I have taken the liberty of composting your answer into $\LaTeX$. If you wish to see the source, right click on the equation above and choose "Show Math As > TeX Commands" – robjohn Feb 21 '13 at 10:17

What you need is the expansion of $(a+b)^4$. This is a special case of the very important Binomial Theorem. We have $$(a+b)^4=a^4+\binom{4}{1}a^3b+\binom{4}{2}a^2b^2+\binom{4}{3}ab^3+b^4.$$ This simplifies to $a^4+4a^3b+6a^2b^2+4ab^3+b^4$.

Put $a=y$, $b=1$, then $a=y$, $b=-1$, and add. There is a pleasant amount of cancellation.

Remarks: $1.$ The symmetrizing move $y=x+4$ is a useful idea.

$2$. For reasons of familiarity, we change the name, and study $(x+1)^4+(x-1)^4$. This function is symmetric about $x=0$. Our function is not $16$ at $x=0$, and by symmetry there are just as many solutions of $(x+1)^4+(x-1)^4=16$ with $x\gt 0$ as there are with $x\lt 0$. So let's see how many there are with $x\gt 0$.

$3.$ The solution $x=1$ is obvious. It is reasonably clear that there are no solutions with $0\lt x\lt 1$. And past $x=1$, our function is increasing. So there is exactly one positive solution, and therefore altogether there are two solutions.

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Multiply: $(y-1)^4-(y+1)^4=(y^4-4y^3+6y^2-4y+1)+(y^4+4y^3+6y^2+4y+1)$. Add.

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