What is the answer to solvefor a,a+b=0a^{2021}+b^{2021} ?

The answer to solvefor a,a+b=0a^{2021}+b^{2021} is a=b^{2021}-b

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solvefor a,a+b=0a^{2021}+b^{2021}

solve for

a, a + b = 0 a^{2021} + b^{2021}

a = b^{2021} - b

Solution steps

a + b = 0 \cdot a^{2021} + b^{2021}

Apply rule

0 \cdot a = 0

a + b = 0 + b^{2021}

0 + b^{2021} = b^{2021}

a + b = b^{2021}

Move

b

to the right side

a = b^{2021} - b

- 3 x - 2 = 6 x

x, 6 x + 9 - 4 y + 4 = - 8 x + 16 + 10 y + 25

a = 2 x - 9

x + 3 (5 x - 1) = 1

f, f = (((i)^{\frac{1}{2}})^{i})^{\frac{1}{2}}

What is the answer to solvefor a,a+b=0a^{2021}+b^{2021} ?
The answer to solvefor a,a+b=0a^{2021}+b^{2021} is a=b^{2021}-b