d Problem: factorise 2 root2 x^3 +3 root3y^3​ | Job Exam Rare Mathematics (JEMrare) - Solved MCQs

# Problem: factorise 2 root2 x^3 +3 root3y^3​

Problem: factorise 2 root2 x^3 +3 root3y^3​

Solution:

The expression can be written as,

$2\sqrt2x^3+3\sqrt3y^3$

This can be written as ,

$=2^1\times 2^{\frac{1}{2}}x^3 +3\times 3^{\frac{1}{2}}y^3$

or

$=2^{1+\frac{1}{2}}x^3+3^{1+\frac{1}{2}}y^3$

or

$=2^{\frac{3}{2}}x^3+3^{\frac{3}{2}}y^3$

Now to factorize it we will give it shape of ​$$a^3+b^3$$​ formula !

So,

$= (2^\frac{1}{2}x)^3+(3^\frac{1}{2}y)^3$

So we know the formula;

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

so,

$(2^\frac{1}{2}x)^3+(3^\frac{1}{2}y)^3=(2^\frac{1}{2}x+3^\frac{1}{2}y){(2^\frac{1}{2}x)^2}-2^\frac{1}{2}x3^\frac{1}{2}y+{(3^\frac{1}{2}y)^2}$

Which can be further simplified as;

$(2^\frac{1}{2}x)^3+(3^\frac{1}{2}y)^3=(2^\frac{1}{2}x+3^\frac{1}{2}y)(2x^2-6^\frac{1}{2}xy+3y^2)$

or

$(2^\frac{1}{2}x)^3+(3^\frac{1}{2}y)^3= (\sqrt2x+\sqrt3y)(2x^2-\sqrt6xy+\3y^2)$