Simplification of algebra.

Question

How to show that $(x^{1/4}-y^{1/4})(x^{3/4}+x^{1/2}y^{1/4}+x^{1/4}y^{1/2}+y^{3/4})=x-y$

Can anyone explain how to solve this question for me? Thanks in advance.

Answer 1

This follows from the fact that $$x^4-y^4=(x^2-y^2)(x^2+y^2)=(x-y)(x+y)(x^2+y^2)=(x-y)(x^3+x^2y+xy^2+y^3)$$ Now just replace $x,y$ with $x^{\frac{1}{4}}$ and $y^{\frac{1}{4}}$.

It is known, in general that $$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\dots+xy^{n-2}+y^{n-1})$$ As can be seen here.

Answer 2

Let $x = u^4$ and $y = v^4$, then the left hand side of the equation simplifies to: $$ (u-v)(u^3 + u^2v + uv^2 + v^3) $$

Now, you can choose to multiply directly: $$ u^4 + \color{red}{u^3v} + \color{green}{u^2v^2} + \color{blue}{uv^3} - \color{red}{vu^3} - \color{green}{v^2u^2} - \color{blue}{v^3u} - v^4 $$ Terms of similar colour cancel out, and you get the result after putting back $x$ and $y$.

Answer 3

Start with $$(x^{1/4}-y^{1/4})(x^{3/4}+x^{1/2}y^{1/4}+x^{1/4}y^{1/2}+y^{3/4}).$$ Then distribute the terms: $$=(x+x^{3/4}y^{1/4}+x^{1/2}y^{1/2}+x^{1/4}y^{3/4})-(x^{3/4}y^{1/4}+x^{1/2}y^{1/2}+x^{1/4}y^{3/4}+y).$$ Regrouping them gives $$(x-y)+(x^{3/4}y^{1/4}-x^{3/4}y^{1/4})+(x^{1/2}y^{1/2}-x^{1/2}y^{1/2})+(x^{1/4}y^{3/4}-x^{1/4}y^{3/4}).$$ The terms cancel out, giving $$(x-y)+0+0+0=x-y.$$

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