Question

x^3-y^3=(x-y)(x^2+xy+y^2​

Answer 1

Step-by-step explanation:

i feel you need to prove this.

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

$\implies \: {x}^{3} - {y}^{3} = {(x - y)}^{3} + 3 {x}^{2} y - 3x {y}^{2}$

$\implies \: {x}^{3} - {y}^{3} = {(x - y)}^{3} + 3xy(x - y)$

$\implies \: {x}^{3} - {y}^{3} = (x - y) \times ( {(x - y)}^{2} + 3xy)$

$\implies \: {x}^{3} - {y}^{3} = (x - y) \times ( {x}^{2} - 2xy + {y}^{2} + 3xy)$

$\implies \: {x}^{3} - {y}^{3} = (x - y) \times ( {x}^{2} + xy + {y}^{2} )$