Maharashtra State BoardSSC (English Medium) 8th Standard
Advertisement Remove all ads

Factorise. Y 3 + 1 8 Y 3 - Mathematics

Sum

Factorise.

`y^3 + 1/(8y^3)`

Advertisement Remove all ads

Solution

It is known that,

\[a^3  +  b^3  = \left( a + b \right)\left( a^2 + b^2 - ab \right)\]

\[\   y^3  + \frac{1}{8 y^3}\] 

\[ =  \left( y \right)^3  +  \left( \frac{1}{2y} \right)^3 \] 

\[ = \left( y + \frac{1}{2y} \right)\left\{ \left( y \right)^2 + \left( \frac{1}{2y} \right)^2 - \left( y \right) \times \left( \frac{1}{2y} \right) \right\}\] 

\[ = \left( y + \frac{1}{2y} \right)\left( y^2 + \frac{1}{4 y^2} - \frac{1}{2} \right)\]

  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

Balbharati Mathematics 8th Standard Maharashtra State Board
Chapter 6 Factorisation of Algebraic expressions
Practice Set 6.2 | Q 6 | Page 31
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×