What are the solutions to the equation solvefor y,x^4+(x-y)^4=0 ?

The solutions to the equation solvefor y,x^4+(x-y)^4=0 are y=-\sqrt[4]{-x^4}+x,y=\sqrt[4]{-x^4}+x

Find the zeros of solvefor y,x^4+(x-y)^4=0

The zeros of solvefor y,x^4+(x-y)^4=0 are y=-\sqrt[4]{-x^4}+x,y=\sqrt[4]{-x^4}+x

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solve for

y, x^{4} + (x - y)^{4} = 0

Solution steps

x^{4} + (x - y)^{4} = 0

Move

x^{4}

to the right side

(x - y)^{4} = - x^{4}

For

(g (x))^{n} = f (a)

, n is even, the solutions are

Solve

The solutions are

s^{3} + 4 s^{2} + 6 s + 5 = 0

4 x^{3} - 2 x^{2} + 5 x = - 7

a^{5} = 3^{5}

x^{3} + 1 3^{2} + 30 x = 0

(- 8 x - 14)^{6} = (- 6 x - 12)^{6}

What are the solutions to the equation solvefor y,x^4+(x-y)^4=0 ?
The solutions to the equation solvefor y,x^4+(x-y)^4=0 are y=-\sqrt[4]{-x^4}+x,y=\sqrt[4]{-x^4}+x
Find the zeros of solvefor y,x^4+(x-y)^4=0
The zeros of solvefor y,x^4+(x-y)^4=0 are y=-\sqrt[4]{-x^4}+x,y=\sqrt[4]{-x^4}+x