Advertisements
Advertisements
प्रश्न
Find the cube of the following binomials expression :
\[4 - \frac{1}{3x}\]
Advertisements
उत्तर
Give `(4- 1/(3x))^3`
We shall use the identity `a^3- b^3 = a^3-b^3 - 3ab(a-b)`
Here `a=4,b=1/(3x)`
By applying in identity we get
`(4- 1/(3x))^3 = (4)^3 - (1/(3x))^3 - 3(4) (1/(3x)) (4-1/(3x))`
`= 4 xx 4xx 4 - (1 xx 1xx1)/(3x xx 3x xx 3x) - 12/(3x) (4-1/(3x))`
` = 64 - 1/(27x^3) - 4/x (4- 1/(3x))`
` = 64 - 1/(27x^3) - (4/x xx 4)-(4/x xx 1/(3x))`
` = 64 - 1/27x^3 - (16/x - 4/(3x^2))`
` = 64 - 1/27x^3 - 16/x + 4/(3x^2)`
Hence cube of the binomial expression of `(4- 1/(3x))^3` is `64 - 1/(27x^3) - 16/x + 4/(3x^2).`
APPEARS IN
संबंधित प्रश्न
Expand the following, using suitable identity:
`[1/4a-1/2b+1]^2`
Evaluate the following using identities:
(2x + y) (2x − y)
Evaluate the following using identities:
117 x 83
If a + b + c = 9 and ab + bc + ca = 23, find the value of a2 + b2 + c2.
Simplify the expression:
`(x + y + z)^2 + (x + y/2 + 2/3)^2 - (x/2 + y/3 + z/4)^2`
If `x - 1/x = 3 + 2sqrt2`, find the value of `x^3 - 1/x^3`
If x = 3 and y = − 1, find the values of the following using in identify:
\[\left( \frac{x}{7} + \frac{y}{3} \right) \left( \frac{x^2}{49} + \frac{y^2}{9} - \frac{xy}{21} \right)\]
(a − b)3 + (b − c)3 + (c − a)3 =
If a + b = 7 and ab = 10; find a - b.
Use the direct method to evaluate :
(4+5x) (4−5x)
Evaluate: 203 × 197
Simplify by using formula :
(5x - 9) (5x + 9)
Simplify by using formula :
(x + y - 3) (x + y + 3)
If x + y = 9, xy = 20
find: x - y
If `"a" - 1/"a" = 10`; find `"a"^2 - 1/"a"^2`
Simplify:
(2x - 4y + 7)(2x + 4y + 7)
Evaluate the following :
1.81 x 1.81 - 1.81 x 2.19 + 2.19 x 2.19
Find the value of x3 – 8y3 – 36xy – 216, when x = 2y + 6
