Question

factorise x^3+3x^2y+3xy^2+y^3

Properly factorise and who ever sent it first I will make him brainly

Answer 1

Answer:

please mark my answer as the brainliest if it helps you mate ❤❤

Step-by-step explanation:

Step  1  :

Equation at the end of step  1  :

(((x3)+((3•(x2))•y))+3xy2)+y3

Step  2  :

Equation at the end of step  2  :

(((x3) + (3x2 • y)) + 3xy2) + y3

Step  3  :

Checking for a perfect cube :

 3.1    Factoring:  x3+3x2y+3xy2+y3 

 .

  x3+3x2y+3xy2+y3  is a perfect cube which means it is the cube of another polynomial 

 In our case, the cubic root of  x3+3x2y+3xy2+y3  is  x+y  

 Factorization is  (x+y)3

Final result :

(x + y)3

Answer 2

Answer:

Most of the explanations assume that the cube of binomial (x + y) is already  embedded in people's brain and the moment you spot the expansion of (x+y)^3, you will be at the solution. That's a wrong approach. Instead, lets

follow this

Step-by-step explanation:

Expand and rearrange the terms and try to find out possible ways to arrive at the simplest binomial power that is a, square of (x+y), which is $x^{2} + 2xy + y^{2}$;

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

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

And, hence the cube.