What is the answer to solvefor A,A^2X^{-1}BA^{-2}=A^5BA^{-3} ?

The answer to solvefor A,A^2X^{-1}BA^{-2}=A^5BA^{-3} is A= 1/(sqrt(X)),A=-1/(sqrt(X));\quad X\ne 0

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solvefor A,A^2X^{-1}BA^{-2}=A^5BA^{-3}

solve for

A, A^{2} X^{- 1} B A^{- 2} = A^{5} B A^{- 3}

A = \frac{1}{X}, A = - \frac{1}{X}; X \neq = 0

Solution steps

A^{2} X^{- 1} B A^{- 2} = A^{5} B A^{- 3}

Refine

\frac{B}{X} = A^{2} B

Multiply both sides by

X

B = A^{2} BX; X \neq = 0

Solve

B = A^{2} BX : A = \frac{1}{X}, A = - \frac{1}{X}; X \neq = 0

A = \frac{1}{X}, A = - \frac{1}{X}; X \neq = 0

- \frac{1}{x ^{2}} = - 1

- 1 - \frac{x}{x - 9} = \frac{1}{x - 9}

0.002 = \frac{c - 52.6}{c} \cdot 0.0035

2 - \frac{3}{x + 4} = \frac{12}{x ^{2} + 4 x}

\frac{1}{2 h} + \frac{5}{h} = \frac{3}{h - 1}

What is the answer to solvefor A,A^2X^{-1}BA^{-2}=A^5BA^{-3} ?
The answer to solvefor A,A^2X^{-1}BA^{-2}=A^5BA^{-3} is A= 1/(sqrt(X)),A=-1/(sqrt(X));\quad X\ne 0