Advertisements
Advertisements
प्रश्न
Find the value of 27x3 + 8y3, if 3x + 2y = 14 and xy = 8
Advertisements
उत्तर
In the given problem, we have to find the value of `27x^3 + 8y^3`
Given `3x + 2y = 14, xy = 8`
On cubing both sides we get,
`(3x+ 2y)^3 = (14)^3`
We shall use identity `(a+b)^3 = a^3 + b^3 + 3ab(a+b)`
`27x^3 + 8y^3 + 3(3x) (2y) (3x+ 2y) = 14 xx 14 xx 14`
`27x^3 + 8y^3 +18(xy)(3x+2y) = 14 xx 14 xx 14`
`27x^3 + 8y^3 + 18(8)(14) = 2744`
`27x^3 + 8y^3 + 2016 = 2744`
` 27x^3 + 8y^3 = 2744 -2016`
`27x^3 +8y^3 = 728`
Hence the value of `27x^3 +8y^3`is 728.
APPEARS IN
संबंधित प्रश्न
Without actually calculating the cubes, find the value of the following:
(28)3 + (–15)3 + (–13)3
Write in the expanded form:
`(a/(bc) + b/(ca) + c/(ab))^2`
Write in the expanded form: (-2x + 3y + 2z)2
If a + b + c = 9 and ab + bc + ca = 23, find the value of a2 + b2 + c2.
Simplify the following expressions:
`(x^2 - x + 1)^2 - (x^2 + x + 1)^2`
If \[x + \frac{1}{x} = 3\], calculate \[x^2 + \frac{1}{x^2}, x^3 + \frac{1}{x^3}\] and \[x^4 + \frac{1}{x^4}\]
If `x^4 + 1/x^4 = 194, "find" x^3 + 1/x^3`
Find the following product:
\[\left( \frac{3}{x} - \frac{5}{y} \right) \left( \frac{9}{x^2} + \frac{25}{y^2} + \frac{15}{xy} \right)\]
Find the following product:
(2ab − 3b − 2c) (4a2 + 9b2 +4c2 + 6 ab − 6 bc + 4ca)
If \[x + \frac{1}{x} = 3\] then \[x^6 + \frac{1}{x^6}\] =
If \[x^3 - \frac{1}{x^3} = 14\],then \[x - \frac{1}{x} =\]
Find the square of : 3a + 7b
Evaluate: (6 − 5xy) (6 + 5xy)
Expand the following:
(a + 4) (a + 7)
Simplify:
(3x + 5y + 2z)(3x - 5y + 2z)
If `x/y + y/x = -1 (x, y ≠ 0)`, the value of x3 – y3 is ______.
Expand the following:
(3a – 5b – c)2
Expand the following:
(3a – 2b)3
Find the value of x3 + y3 – 12xy + 64, when x + y = – 4
