Question

Find the value of x3 + y3 + z3 – 3xyz if x2 + y2 + z2 = 83 and x + y + z = 15 verified answer

Answer 1

Answer:

68

Step-by-step explanation:

We know that x2+y2+z2=83----------------(1)

                       x+y+z=15-----------------------(2)

Here,

x3+y3+z3-3xyz=(x2+y2+z2-xy-yz-xz)

        [SINCE a3+b3+c3-3abc=(a2+b2+c2_ab-bc-ca)]

x3+y3+z3-3xyz={x2+y2+z2-(xy+yz+xz)}

x3+y3+z3-3xyz={83-(15)}

x3+y3+z3-3xyz=68

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Answer 2

Step by step explaination :

According to the identity,

(x+y+z)^2=x^2+y^2+z^3+2(xy+yz+zx)

Substituting values:

(15)^2=83+2(xy+yz+zx)

225-83=2()

142=2(xy+yz+zx)

xy+yz+zx=71

According to the problem,

x3 + y3 + z3 – 3xyz=(x+y+z)(x2 + y2 + z2 -(xy+yz+zx))

Substituting values:

15*12=180

ANSWER=180

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