Solve the following system of equations: x3+y3=7,x3y3=−8
Given system of equations:
x3+y3=7 ⋯(1) and x3y3=−8 ⋯(2)
Equation (2) can be written as
(xy)3=(−2)3
i.e. xy=−2
i.e. x=−2y ⋯(3)
In view of equation (3), equation (1) becomes
(−2y)3+y3=7
i.e. −8y3+y3=7
i.e. y6−7y3−8=0 ⋯(4)
Let y3=p ⋯(5)
The equation (4) becomes
p2−7p−8=0
i.e. (p−8)(p+1)=0
i.e. p=8 or p=−1
Substituting the value of p in equation (5) gives
when p=8, y3=8 i.e. y=2
and when p=−1, y3=−1 i.e. y=−1
Substituting the value of y in equation (3) gives
when y=2, x=−22=−1
and when y=−1, x=−2−1=2
∴ The solution to the given system of equations is \;\; \left(x, y\right) = \left\{\left(-1, 2\right), \; \left(2, -1\right) \right\}