Advertisements
Advertisements
Question
If 2x+3y = 13 and xy = 6, find the value of 8x3 + 27y3
Advertisements
Solution
In the given problem, we have to find the value of `8x^3 + 27y^3`
Given`2x + 3y = 13, xy = 6`
In order to find `8x^3 + 27y^3`we are using identity `(a+b )^3 = a^3 + b^3 + 3ab(a+b)`
`(2x + 3y )^3 = (13)^3`
`8x^3 + 27 y^3 + 3 (2x)(3y)(2x+ 3y)= 2197`
` 8x^3 + 27y^3 + 18xy (2x+ 3y) = 2197`
Here putting, `2x + 3y = 13, xy = 6`
`8x^3 + 27y^3 + 18 xx 6 xx 13 = 2197`
` 8x^3 + 27y^3 + 1404 = 2197`
` 8x^3 + 27y^3 = 2197 - 1404`
`8x^3+ 27y^3 = 793`
Hence the value of `8x^3 + 27y^3` is 793.
APPEARS IN
RELATED QUESTIONS
Use suitable identity to find the following product:
(3x + 4) (3x – 5)
Expand the following, using suitable identity:
(–2x + 3y + 2z)2
Write the following cube in expanded form:
(2a – 3b)3
Evaluate the following using identities:
(2x + y) (2x − y)
If `x^2 + 1/x^2 = 66`, find the value of `x - 1/x`
Find the value of 4x2 + y2 + 25z2 + 4xy − 10yz − 20zx when x = 4, y = 3 and z = 2.
If a − b = 4 and ab = 21, find the value of a3 −b3
Evaluate the following:
(98)3
Evaluate of the following:
933 − 1073
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 + \[\frac{1}{x}\] = then find the value of \[x^2 + \frac{1}{x^2}\].
Evaluate: (9 − y) (7 + y)
Evaluate: `(4/7"a"+3/4"b")(4/7"a"-3/4"b")`
Find the squares of the following:
`(7x)/(9y) - (9y)/(7x)`
Simplify by using formula :
(x + y - 3) (x + y + 3)
Evaluate the following without multiplying:
(999)2
If `x + (1)/x = 3`; find `x^2 + (1)/x^2`
If p2 + q2 + r2 = 82 and pq + qr + pr = 18; find p + q + r.
Expand the following:
(4a – b + 2c)2
Prove that (a + b + c)3 – a3 – b3 – c3 = 3(a + b)(b + c)(c + a).
