The number of solutions of the equations x+y+ | vedclass

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(A) Given equations are:$x+y+z=12$ ... $(i)$$x^2+y^2+z^2=50$ ... $(ii)$$x^3+y^3+z^3=216$ ... $(iii)$Using the identity $(x+y+z)^2 = x^2+y^2+z^2+2(xy+y

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$6$

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(A) Given equations are:$x+y+z=12$ ... $(i)$$x^2+y^2+z^2=50$ ... $(ii)$$x^3+y^3+z^3=216$ ... $(iii)$Using the identity $(x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx)$:$12^2 = 50 + 2(xy+yz+zx)$$144 - 50 = 2(xy+yz+zx)$$xy+yz+zx = 47$Using the identity $x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-(xy+yz+zx))$:$216 - 3xyz = 12(50 - 47)$$216 - 3xyz = 12(3) = 36$$3xyz = 180 \Rightarrow xyz = 60$Now,$x, y, z$ are roots of the cubic equation $t^3 - (x+y+z)t^2 + (xy+yz+zx)t - xyz = 0$:$t^3 - 12t^2 + 47t - 60 = 0$Testing roots,for $t=3$: $27 - 12(9) + 47(3) - 60 = 27 - 108 + 141 - 60 = 0$. So $(t-3)$ is a factor.$(t-3)(t^2-9t+20) = 0$$(t-3)(t-4)(t-5) = 0$The roots are $3, 4, 5$.Since $x, y, z$ can be any permutation of ${3, 4, 5}$,the number of solutions is $3! = 6$.

The number of solutions of the equations $x+y+z=12$,$x^2+y^2+z^2=50$,and $x^3+y^3+z^3=216$ is

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