Advertisements
Advertisements
प्रश्न
If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.
उत्तर
Since, x + y + z = 0
= x + y = −z(x + y)3 = (−z)3
= x3 + y3 + 3xy(x + y) = (−z)3
= x3 + y3 + 3xy(−z) = −z3 ...[∵ x + y = −z]
= x3 + y3 − 3xyz = (−z)3
= x3 + y3 + z3 = 3xyz
Hence, if x + y + z = 0, then
x3 + y3 + z3 = 3xyz
APPEARS IN
संबंधित प्रश्न
Expand the following, using suitable identity:
(–2x + 5y – 3z)2
Factorise the following:
27y3 + 125z3
What are the possible expressions for the dimensions of the cuboids whose volume is given below?
| Volume : 3x2 – 12x |
If \[x - \frac{1}{x} = - 1\] find the value of \[x^2 + \frac{1}{x^2}\]
If \[x + \frac{1}{x} = 5\], find the value of \[x^3 + \frac{1}{x^3}\]
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:
(3x + 2y) (9x2 − 6xy + 4y2)
If a + b = 10 and ab = 16, find the value of a2 − ab + b2 and a2 + ab + b2
Find the following product:
(3x + 2y + 2z) (9x2 + 4y2 + 4z2 − 6xy − 4yz − 6zx)
If x + \[\frac{1}{x}\] = then find the value of \[x^2 + \frac{1}{x^2}\].
Mark the correct alternative in each of the following:
If \[x + \frac{1}{x} = 5\] then \[x^2 + \frac{1}{x^2} = \]
If \[3x + \frac{2}{x} = 7\] , then \[\left( 9 x^2 - \frac{4}{x^2} \right) =\]
Use the direct method to evaluate the following products :
(5a + 16) (3a – 7)
Use the direct method to evaluate :
`("z"-2/3)("z"+2/3)`
Expand the following:
(x - 5) (x - 4)
If p + q = 8 and p - q = 4, find:
pq
If p2 + q2 + r2 = 82 and pq + qr + pr = 18; find p + q + r.
Simplify:
(2x - 4y + 7)(2x + 4y + 7)
Without actually calculating the cubes, find the value of:
(0.2)3 – (0.3)3 + (0.1)3
