If x=y+z then find the value of x3-y3-z3-3xyz
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Answered by
8
x = y + z
=> x + (-y) +(-z) = 0
[ By using the identity,
If a+b+c = 0
Then a^3 + b^3 + c^3 = 3abc]
x^3 - y^3 - z^3 = 3(x) (-y) (-z)
x^3 - y^3 - z^3 = 3xyz -------(1)
Now,
x^3 - y^3 - z^3 -3xyz = 3xyz - 3xyz = 0
=> x + (-y) +(-z) = 0
[ By using the identity,
If a+b+c = 0
Then a^3 + b^3 + c^3 = 3abc]
x^3 - y^3 - z^3 = 3(x) (-y) (-z)
x^3 - y^3 - z^3 = 3xyz -------(1)
Now,
x^3 - y^3 - z^3 -3xyz = 3xyz - 3xyz = 0
Answered by
4
x

x

x
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