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
Evaluate the following product without multiplying directly:
104 × 96
Factorise:
`2x^2 + y^2 + 8z^2 - 2sqrt2xy + 4sqrt2yz - 8xz`
What are the possible expressions for the dimensions of the cuboids whose volume is given below?
| Volume : 12ky2 + 8ky – 20k |
Simplify the following: 175 x 175 x 2 x 175 x 25 x 25 x 25
Simplify the following products:
`(x/2 - 2/5)(2/5 - x/2) - x^2 + 2x`
Write in the expanded form:
`(2 + x - 2y)^2`
If a − b = 4 and ab = 21, find the value of a3 −b3
If \[x + \frac{1}{x} = 3\], calculate \[x^2 + \frac{1}{x^2}, x^3 + \frac{1}{x^3}\] and \[x^4 + \frac{1}{x^4}\]
Find the following product:
If x = 3 and y = − 1, find the values of the following using in identify:
\[\left( \frac{x}{7} + \frac{y}{3} \right) \left( \frac{x^2}{49} + \frac{y^2}{9} - \frac{xy}{21} \right)\]
If x = −2 and y = 1, by using an identity find the value of the following
75 × 75 + 2 × 75 × 25 + 25 × 25 is equal to
If \[x^4 + \frac{1}{x^4} = 623\] then \[x + \frac{1}{x} =\]
Use the direct method to evaluate the following products:
(a – 8) (a + 2)
Use the direct method to evaluate the following products :
(b – 3) (b – 5)
Use the direct method to evaluate the following products:
(5a + 16) (3a – 7)
Use the direct method to evaluate :
(2+a) (2−a)
Simplify by using formula :
(x + y - 3) (x + y + 3)
Evaluate, using (a + b)(a - b)= a2 - b2.
4.9 x 5.1
Simplify:
(1 + x)(1 - x)(1 - x + x2)(1 + x + x2)
