Advertisements
Advertisements
प्रश्न
Find the following product:
Advertisements
उत्तर
Given `(1-x)(1 + x + x^2)`
We shall use the identity `(a-b)(a^2+ ab + b^2) = a^3 - b^3`
We can rearrange the `(1 - x) (1+ x + x^2)`as
` = (1- x) [(1)^2 + (1)(x)+ (x)^2]`
` = (1)^3 - (x)^3`
` = (1) xx (1) xx (1) - (x) xx (x) xx (x)`
` = 1=x^3`
Hence the Product value of `(1-x)(1+x + x^2)`is `1-x^3`.
APPEARS IN
संबंधित प्रश्न
Use suitable identity to find the following product:
(x + 8) (x – 10)
Evaluate the following product without multiplying directly:
103 × 107
Factorise:
4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
Factorise the following:
`27p^3-1/216-9/2p^2+1/4p`
If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.
Evaluate the following using identities:
117 x 83
Simplify the expression:
`(x + y + z)^2 + (x + y/2 + 2/3)^2 - (x/2 + y/3 + z/4)^2`
Evaluate of the following:
(598)3
If x = 3 and y = − 1, find the values of the following using in identify:
(9y2 − 4x2) (81y4 +36x2y2 + 16x4)
If x = 3 and y = − 1, find the values of the following using in identify:
\[\left( \frac{x}{y} - \frac{y}{3} \right) \frac{x^2}{16} + \frac{xy}{12} + \frac{y^2}{9}\]
If x + y + z = 8 and xy +yz +zx = 20, find the value of x3 + y3 + z3 −3xyz
If \[a^2 + \frac{1}{a^2} = 102\] , find the value of \[a - \frac{1}{a}\].
If a + b + c = 0, then \[\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab} =\]
Use identities to evaluate : (97)2
If a2 - 5a - 1 = 0 and a ≠ 0 ; find:
- `a - 1/a`
- `a + 1/a`
- `a^2 - 1/a^2`
Find the squares of the following:
(2a + 3b - 4c)
If `"a" + (1)/"a" = 2`, then show that `"a"^2 + (1)/"a"^2 = "a"^3 + (1)/"a"^3 = "a"^4 + (1)/"a"^4`
Using suitable identity, evaluate the following:
101 × 102
Expand the following:
`(1/x + y/3)^3`
Give possible expressions for the length and breadth of the rectangle whose area is given by 4a2 + 4a – 3.
