Question

prove that (cube root of 3x -y) is a factor of p(x,y)=405x^6 + 45y^6-270x^3y^3​

Answer 1

Given: $p(x,y)=405x^{6}+45y^{6}-270x^{3}y^{3}$

To prove: $(\sqrt[3]{3}x-y)$ a factor of $p(x,y)$

Proof:

Now, $405x^{6}+45y^{6}-270x^{3}y^{3}$

$=45\:(9x^{6}+y^{6}-6x^{3}y^{3})$

$=45\:[(3x^{3})^{2}+(y^{3})^{2}-2\times 3x^{3}\times y^{3}$

$=45\:(3x^{3}-y^{3})^{2}$

$=45\:\{(\sqrt[3]{3}x)^{3}-(y)^{3}\}^{2}$

$=45\:[(\sqrt[3]{3}x-y)\:\{(\sqrt[3]{3}x)^{2}+\sqrt[3]{3}xy+y^{2}\}]^{2}$

$\longrightarrow$ This shows that $(\sqrt[3]{3}x-y)$ is a factor of $p(x,y)$.

Hence proved.

Formula used:

• $a^{2}-2ab+b^{2}=(a-b)^{2}$

• $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$