Question

If (x+y+z)=9 & x^2+y^2+z^2=35. Find the value of x^3+y^3+z^3-3xyz

Answer 1

we know that (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)

where x=a

y=b

and z=c

so we have given that x+y+z=9

and x^2+y^2+z^2=35

9^2=35+2(xy+xz+yz)

81=35+2(xy+yz+zx)

81-35/2=xy+yz+zx

23=xy+yz+zx

now we know that x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^-(xy+yz+zx)

now by substituting the value we get

9(35-23)

9(12)

= 108

your answer is 108

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