Advertisements
Advertisements
प्रश्न
If a + b = 11 and a2 + b2 = 65; find a3 + b3.
Advertisements
उत्तर
a + b = 11 and a2 + b2 = 65
Now, (a+b)2 = a2 + b2 + 2ab
⇒ (11)2 = 65 + 2ab
⇒ 121 = 65 + 2ab
⇒ 2ab = 56
⇒ ab = 28
a3 + b3 = ( a + b )( a2 - ab + b2)
= (11)(65 - 28)
= 11 x 37
= 407
APPEARS IN
संबंधित प्रश्न
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`
( 2x + 3y )( 4x2 + 6xy + 9y2 )
Find : (a + b)(a + b)(a + b)
Prove that : x2+ y2 + z2 - xy - yz - zx is always positive.
If x = 3 + 2√2, find :
(i) `1/x`
(ii) `x - 1/x`
(iii) `( x - 1/x )^3`
(iv) `x^3 - 1/x^3`
If x + 5y = 10; find the value of x3 + 125y3 + 150xy − 1000.
Using suitable identity, evaluate (104)3
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]`
