Advertisements
Advertisements
Question
Using suitable identity, evaluate (104)3
Advertisements
Solution
Using identity: (a ± b)3 = a3 ± b3 ± 3ab(a ± b)
(104)3 = (100 + 4)3
= (100)3 + (4)3 + 3 × 100 × 4(100 + 4)
= 1000000 + 64 + 1200 × 104
= 1000000 + 64 + 124800
= 1124864
APPEARS IN
RELATED QUESTIONS
Simplify : ( x + 6 )( x + 4 )( x - 2 )
Simplify : ( x - 6 )( x - 4 )( x - 2 )
Simplify using following identity : `( a +- b )(a^2 +- ab + b^2) = a^3 +- b^3`
`( 3x - 5/x )( 9x^2 + 15 + 25/x^2)`
Find : (a + b)(a + b)(a + b)
Prove that : x2+ y2 + z2 - xy - yz - zx is always positive.
If a + b = 11 and a2 + b2 = 65; find a3 + b3.
If a − 2b + 3c = 0; state the value of a3 − 8b3 + 27c3.
Using suitable identity, evaluate (97)3
Simplify :
`[(x^2 - y^2)^3 + (y^2 - z^2)^3 + (z^2 - x^2)^3]/[(x - y)^3 + (y - z)^3 + (z - x)^3]`
Evaluate :
`[0.8 xx 0.8 xx 0.8 + 0.5 xx 0.5 xx 0.5]/[0.8 xx 0.8 - 0.8 xx 0.5 + 0.5 xx .5]`
