By Fermat's Last Theorem, the above equation remains true if at least one of the three expressions equal 0.
Case 1: x^2+3x-4=0
The above equation yields x=1, -4
Substituting x=1, we have:
LHS = RHS of the given equation.
Substituting x=-4, we have:
LHS = 55^3
RHS = 55^3, so that x=-4 is a valid solution..
Case 2: 2x^2 -5x+3=0
x= 1, 3/2
We have already checked, and found x=1 to be a valid solution.
Substituting x=3/2, we have:
LHS =(11/4)^3
RHS = (11/4)^3
Therefore, x= 3/2 is a valid solution as LHS =RHS
Case 3: 3x^2-2x-1=0
x=1, -1/3
Substituting x=-1/3, we have:
LHS = (-44/9)^3 + (44/9)^3 =0
RHS = 0
Therefore, x = -1/3 is also a valid solution.
Consequently, x=1, 3/2, -1/3, -4 are the required solutions to the given equation.
Edited on May 12, 2024, 8:50 am
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