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Here they give a classic decomposition but I'm not sure if there's a quick method to arrive at the right hand side from the left, or if I should just memorize this decmomposition.

Thanks

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john fowles
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4 Answers4

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If you consider this as a polynomial in $x$, then $x=-(a+b)$ is a root of left hand side polynomial. Then factorize.

Another approach could be to use $\omega$, the complex cube root of unity. $$LHS=(x+a+b)(x+a\omega+b\omega^2)(x+a\omega^2+b\omega)$$

Another approach could be to add and subtract $3x^2a+3xa^2$ term and then factor. \begin{align}x^3+a^3+b^3-3xab & = x^3+a^3+3x^2a+3xa^2+b^3-3xab-3x^2a-3xa^2 \\ & =(x+a)^3+b^3-3xa(x+a+b)\\ &= (x+a+b)((x+a)^2+b^2-(x+a)b)-3xa(x+a+b) \\ \end{align}

Anurag A
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  • At the moment I studying complex roots of unity, so I would like to understand your approach using that method. So are you assuming the equality we want to prove holds? Because it seems to me that you assumed the equality in order to know that LHS has $(x+a+b)$ as a factor, unless something else told you that it's a factor. – john fowles Aug 03 '18 at 13:59
  • @johnfowles if you substitute $x=-(a+b)$ in the LHS, you get $0$. This shows $x+a+b$ is a factor. Since LHS is symmetric in $x,a,b$ so this is a natural guess for a linear factor. – Anurag A Aug 03 '18 at 18:22
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A straightforward way (maybe not that quick, but easy) is the following. Write $$ a^3 + b^3 + x^3 -3abx = (a + b + x)^3 - 3(a^2b + a^2 x + ab^2 + ax^2 + b^2x + bx^2 + 3abx) $$ and $$ a^2b + a^2 x + ab^2 + ax^2 + b^2x + bx^2 + 3abx = (a + b + x)(ab + bx + ax) $$ concluding with the final result.

A thing to remember is to use the expression $(a + b + x)^3$ and the rest is just technique!

tortue
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$$x^3+a^3+b^3-3xab=x^3+a^3+b^3+3a^2b+3ab^2-3xab-3a^2b-3ab^2=$$ $$=x^3+(a+b)^3-3ab(x+a+b)=(x+a+b)(x^2-x(a+b)+(a_+b)^2-3ab)=$$ $$=(x+a+b)(x^2+a^2+b^2-xa-xb-ab).$$

Michael Rozenberg
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$$x^3+a^3+b^3-3abx$$

$$=x^3+(a+b)^3-3ab(a+b)-3abx$$

$$=\underbrace{(x+a+b)}\{x^2-x(a+b)+(a+b)^2\}-3ab\underbrace{(x+a+b)}$$

$$=?$$

lab bhattacharjee
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