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