Question

factorise 2 root 2 x cube + 3 root 3 y cube​

Answer 1

heya mate ♥️

2√2x³+3√3y³

➜ (√2x)³+(√3y)³

➜ (√2x+√3y)(2x²+√6xy+3y²)

hope it's help you ☺️

Answer 2

Answer:

The factorization of given expression is $2\sqrt{2}x^{3}+3\sqrt{3}y^{3}=(2\sqrt{2}+3\sqrt{3})(35-6\sqrt{6})$

Step-by-step explanation:

The given expression which we are required to factorize is

$2\sqrt{2}x^{3}+3\sqrt{3}y^{3}$

To factorize it we shall covert the given expression in form of $a^{3}+b^{3}$ this can be done as shown below

$2\sqrt{2}x^{3}+3\sqrt{3}y^{3}=\sqrt{2} .\sqrt{2} .\sqrt{2}x^{3}+ \sqrt{3} .\sqrt{3} .\sqrt{3} y^{3}\\=(2\sqrt{2}x)^{3}+(3\sqrt{3}y)^{3}$

We have an algebraic identity for sum of cube of two numbers which is

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

In our case, $a=2\sqrt{2},b=3\sqrt{3}$ substituting these values we get

$(2\sqrt{2}x)^{3}+(3\sqrt{3}y)^{3}=(2\sqrt{2}+3\sqrt{3})[(2\sqrt{2} )^{2}+(3\sqrt{3} )^{2}-(2\sqrt{2})(3\sqrt{3} )]\\=(2\sqrt{2}+3\sqrt{3})(8+27-6\sqrt{6})\\=(2\sqrt{2}+3\sqrt{3})(35-6\sqrt{6})$

Hence,the factorization of given expression is obtained as

$2\sqrt{2}x^{3}+3\sqrt{3}y^{3}=(2\sqrt{2}+3\sqrt{3})(35-6\sqrt{6})$

#SPJ2