71Fine. But i guess there should be another neat solution also.
$\textbf{Solve system of equations:}\\\\ $\mathbf{x+y-z=7}$\\\\ $\mathbf{x^2+y^2-z^2=37}$\\\\ $\mathbf{x^3+y^3-z^3=1}$
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2 Answers
36maybe lengthy ... but i think i can try my hand at this... :P
x + y = 7 + z --- (i)
(x + y)2 - 2xy = 37 + z2
=> (7 + z)2 - 2xy = 37 + z2
=> 2xy = 49 + 14z + z2 - 37 - z2 => 2xy = 12 + 14z
=> xy = 6 + 7z ----- (ii)
Again, x3 + y3 = 1 + z3
=> (x + y)(x2 + y2 - xy) = (1 + z)(1 + z2 - z)
=> (7 + z)(37 + z2 - 6 - 7z) = (1 + z)(1 + z2 - z)
=> (7 + z)(31 + z2 - 7z) = (1 + z)(1 + z2 - z)
=> 217 + 7z2 - 49z + 31z + z3 - 7z2 = 1 + z2 - z + z + z3 - z2
=> 217 - 18z = 1 => 18z = 216 => z = 12
Thus, z = 12
Now its easy to find x and y....!!
71