Advertisements
Advertisements
Question
Factorize the following expressions:
y3 +125
Advertisements
Solution
= y3 + 53
= ( y + 5)( y2 - 5y + 52 ) [ ∵ a3 + b3 = (a + b)(a2 - ab + b2 )]
= ( y + 5)( y2 - 5y + 25)
∴ y3 +125 = ( y + 5)( y2 - 5y + 25)
APPEARS IN
RELATED QUESTIONS
Factorize `x^2 - 2sqrt2x - 30`
Factorize the following expressions:
32a3 + 108b3
Factorize the following expressions:
`8x^2y^3 - x^5`
Simplify `(155 xx 155 xx 155 - 55 xx 55 xx 55)/(155 xx 155 + 155 xx 55 + 55 xx 55)`
Factorize `8/27 x^3 + 1 + 4/3 x^2 + 2x`
If 3x = a + b + c, then the value of (x − a)3 + (x −b)3 + (x − c)3 − 3(x − a) (x − b) (x −c) is
Separate monomials, binomials, trinomials and polynomials from the following algebraic expressions :
8 − 3x, xy2, 3y2 − 5y + 8, 9x − 3x2 + 15x3 − 7,
3x × 5y, 3x ÷ 5y, 2y ÷ 7 + 3x − 7 and 4 − ax2 + bx + y
Multiply: (2x + 3y)(2x + 3y)
Divide: 35a3 + 3a2b - 2ab2 by 5a - b
Divide: m3 − 4m2 + m + 6 by m2 − m − 2
