Advertisements
Advertisements
Question
Find the following product:
(4x − 3y + 2z) (16x2 + 9y2 + 4z2 + 12xy + 6yz − 8zx)
Advertisements
Solution
In the given problem, we have to find Product of equations
Given (4x − 3y + 2z) (16x2 + 9y2 + 4z2 + 12xy + 6yz − 8zx)
We shall use the identity
`x^3 + y^3 + z^3 - 3xyz = (x+ y +z)(x^2 + y^2 + z^2 - xy - yz - zx)`
` = (4x)^3 + (3y)^3 + (2z)^3 -3 (4x)(3y)(2z)`
` = (4x) xx (4x) xx (4x) +(-3y) xx (-3y) xx (-3y) + (2z) xx (2z) xx (2z) -3 (4x) (-3y)(2z)`
` = 64x^3 - 27y^3 + 8z^3 + 72 xyz`
Hence the product of (4x − 3y + 2z) (16x2 + 9y2 + 4z2 + 12xy + 6yz − 8zx) is `64x^2 - 27y^3 + 8z^3 + 72xyz`
APPEARS IN
RELATED QUESTIONS
Write the following cube in expanded form:
(2a – 3b)3
If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.
Evaluate the following using identities:
`(2x+ 1/x)^2`
Simplify the following: 175 x 175 x 2 x 175 x 25 x 25 x 25
Write in the expanded form:
`(2 + x - 2y)^2`
Simplify `(x^2 + y^2 - z)^2 - (x^2 - y^2 + z^2)^2`
Simplify (2x + p - c)2 - (2x - p + c)2
If 3x − 2y = 11 and xy = 12, find the value of 27x3 − 8y3
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 \[x^4 + \frac{1}{x^4} = 119\] , find the value of \[x^3 - \frac{1}{x^3}\]
Find the following product:
\[\left( \frac{3}{x} - \frac{5}{y} \right) \left( \frac{9}{x^2} + \frac{25}{y^2} + \frac{15}{xy} \right)\]
If x = 3 and y = − 1, find the values of the following using in identify:
\[\left( \frac{3}{x} - \frac{x}{3} \right) \left( \frac{x^2}{9} + \frac{9}{x^2} + 1 \right)\]
Find the square of : 3a + 7b
If a2 - 3a + 1 = 0, and a ≠ 0; find:
- `a + 1/a`
- `a^2 + 1/a^2`
If m - n = 0.9 and mn = 0.36, find:
m + n
If m - n = 0.9 and mn = 0.36, find:
m2 - n2.
If a + b + c = 0, then a3 + b3 + c3 is equal to ______.
Give possible expressions for the length and breadth of the rectangle whose area is given by 4a2 + 4a – 3.
Multiply x2 + 4y2 + z2 + 2xy + xz – 2yz by (–z + x – 2y).
