if X= 1,y=2 and z=3,find the value of x³+y³+,z³-3xyz
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Answer:
150
Step-by-step explanation:
According to identity, we know that
(x+y+z)(x²+y²+z²-(xy+yz+zx))=x³+y³+z³-3xyz. Let this be equation 1.
But we don't know x²+y²+z².
We can find it by using
(x+y+z)²=x²+y²+z²+2(xy+yz+zx)
Substituting known values, we get
15²=x²+y²+z²+2(71)
x²+y²+z²=225-142
=83
Substituting known values and x²+y²+z² in equation 1, we get
15*(83-71)=x³+y³+z³-3(10)
x³+y³+z³=180-30
=150.
Thus, 150 is the required answer.
hope this helps
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