Question

factorise 2root 2 x cube + 3 root 3 y cube + 5 root 5 minus 3 root 30 xy

Answer 1

can u give in equation form?

Answer 2

The factorisation of $2\sqrt{2}x^3+3\sqrt{3}y^{3}+5\sqrt{5}-3\sqrt{30}xy$ $=(\sqrt{2}x+\sqrt{3}y)(2x^2-\sqrt{6}xy+3y^2) +\sqrt{5}(5-3\sqrt{6}xy)$

Step-by-step explanation:

We have,

$2\sqrt{2}x^3+3\sqrt{3}y^{3}+5\sqrt{5}-3\sqrt{30}xy$

To find, the factorisation of $2\sqrt{2}x^3+3\sqrt{3}y^{3}+5\sqrt{5}-3\sqrt{30}xy=?$

∴ $2\sqrt{2}x^3+3\sqrt{3}y^{3}+5\sqrt{5}-3\sqrt{30}xy$

$=(\sqrt{2}x)^3+(\sqrt{3}y)^{3}+5\sqrt{5}-3\sqrt{6\times 5}xy$

Using the algebraic identity,

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

$=(\sqrt{2}+(\sqrt{3}y)[(\sqrt{2}x)^2-\sqrt{2}x+\sqrt{3}y(\sqrt{3}y)^{2}]+5\sqrt{5}-3\sqrt{5}\sqrt{6}xy$

$=(\sqrt{2}x+\sqrt{3}y)(2x^2-\sqrt{6}xy+3y^2) +\sqrt{5}(5-3\sqrt{6}xy)$

The factorisation of $2\sqrt{2}x^3+3\sqrt{3}y^{3}+5\sqrt{5}-3\sqrt{30}xy$ $=(\sqrt{2}x+\sqrt{3}y)(2x^2-\sqrt{6}xy+3y^2) +\sqrt{5}(5-3\sqrt{6}xy)$

Hence, the factorisation of $2\sqrt{2}x^3+3\sqrt{3}y^{3}+5\sqrt{5}-3\sqrt{30}xy$ $=(\sqrt{2}x+\sqrt{3}y)(2x^2-\sqrt{6}xy+3y^2) +\sqrt{5}(5-3\sqrt{6}xy)$