Advertisements
Advertisements
प्रश्न
Find the cube of the following binomials expression :
\[\frac{1}{x} + \frac{y}{3}\]
Advertisements
उत्तर
In the given problem, we have to find cube of the binomial expressions
Given `(1/x+y/3)^3`
We shall use the identity `(a+b)^3 = a^3+b^3+3ab(a+b)`
Here `a=1/x ,b=y/3`
By applying the identity we get
`(1/x+y/x)^3 = (1/x)^3 + (y/3)^3+3 (1/x)(y/3)(1/x+y/3)`
` = 1/x^3 +y^3/27+3 xx 1/x xx y/3 (1/x +y/3)`
` = 1/x^3 +y^3/27+ y/x (1/x +y/3)`
` = 1/x^3 +y^3/27+ y/x xx 1/x +y/x xxy/3`
` = 1/x^3 +y^3/27+ y/x^2+y/(3x)`
Hence cube of the binomial expression `1/x+y/3` is `1/x^3 + y^3/27 +y/x^2+y^2/(3x)`
APPEARS IN
संबंधित प्रश्न
Expand the following, using suitable identity:
(x + 2y + 4z)2
Verify that `x^3+y^3+z^3-3xyz=1/2(x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2]`
Evaluate the following using identities:
(2x + y) (2x − y)
Write in the expanded form:
`(2 + x - 2y)^2`
Simplify the expression:
`(x + y + z)^2 + (x + y/2 + 2/3)^2 - (x/2 + y/3 + z/4)^2`
If \[x^2 + \frac{1}{x^2}\], find the value of \[x^3 - \frac{1}{x^3}\]
Find the value of 64x3 − 125z3, if 4x − 5z = 16 and xz = 12.
If a + b = 6 and ab = 20, find the value of a3 − b3
Find the following product:
(2ab − 3b − 2c) (4a2 + 9b2 +4c2 + 6 ab − 6 bc + 4ca)
If \[x + \frac{1}{x} = 2\], then \[x^3 + \frac{1}{x^3} =\]
If a2 + b2 + c2 − ab − bc − ca =0, then
Evalute : `((2x)/7 - (7y)/4)^2`
If a - b = 0.9 and ab = 0.36; find:
(i) a + b
(ii) a2 - b2.
Evaluate: (4 − ab) (8 + ab)
Evaluate: 203 × 197
Simplify by using formula :
(2x + 3y) (2x - 3y)
If `x + (1)/x = 3`; find `x^4 + (1)/x^4`
If m - n = 0.9 and mn = 0.36, find:
m2 - n2.
If x + y = 1 and xy = -12; find:
x2 - y2.
Find the value of x3 – 8y3 – 36xy – 216, when x = 2y + 6
