In the verge of coronavirus pandemic, we are providing FREE access to our entire Online Curriculum to ensure Learning Doesn't STOP!

Ex.2.5 Q9 Polynomials Solution - NCERT Maths Class 9

Go back to  'Ex.2.5'

Question

Verify:

(i) \(\begin{align}\left(x^{3}+y^{3}\right)=(x+y)\left(x^{2}-x y+y^{2}\right) \end{align}\)

(ii) \(\begin{align}\left(x^{3}-y^{3}\right)=(x-y)\left(x^{2}+x y+y^{2}\right)\end{align}\)

   

 Video Solution
Polynomials
Ex 2.5 | Question 9

Text Solution

  

Steps:

(i) \(\begin{align}\left(x^{3}+y^{3}\right)=(x+y)\left(x^{2}-x y+y^{2}\right) \end{align}\)

\[\begin{align}&(x\!+\!y) \left(x^{2}\!-\!x y\!+\!y^{2}\right) \\&= x\left(x^{2}\!-\!x y\!+\!y^{2}\right)\!+\!y\left(x^{2}\!-\!x y\!+\!y^{2}\right) \\&\!=\! x^{3}\!-\!x^{2} y\!+\!x y^{2}\!+\!x^{2} y\!-\!x y^{2}\!+\!y^{3} \\&=\!x^{3}\!+\!y^{3} \end{align}\]

(ii) \(\begin{align}\left(x^{3}-y^{3}\right)=(x-y)\left(x^{2}+x y+y^{2}\right)\end{align}\)

\[\begin{align}&{(x\!-\!y)\left(x^{2}\!+\!x y\!+\!y^{2}\right)} \\ &\!=\!x\left(x^{2}\!+\!x y\!+\!y^{2}\right)\!-\!y\left(x^{2}\!+\!x y\!+\!y^{2}\right) \\ &=x^{3}\!+\!x^{2} y\!+\!x y^{2}\!-\!x^{2} y\!-\!1x y^{2}-\!y^{3} \\ &=x^{3}-y^{3} \end{align}\]

 Video Solution
Polynomials
Ex 2.5 | Question 9