Question

Simplify the expression. 8x^-10 y^'6 -2x^2y^-8 Write your answer without negative exponents.​

Answer 1

Answer:

$8x^{-10}y^6 - 2x^2y^{-8} = \frac{8y^{14} - 2x^{12}}{x^{10}y^8}$

Step-by-step explanation:

Given

$8x^{-10}y^6 - 2x^2y^{-8}$

Required

Simplify

Rewrite as:

$8x^{-10}y^6 - 2x^2y^{-8} = \frac{8y^6}{x^{10}} - \frac{2x^2}{y^8}$

Take LCM

$8x^{-10}y^6 - 2x^2y^{-8} = \frac{8y^6*y^8 - 2x^2 * x^{10}}{x^{10}y^8}$

Apply law of indices

$8x^{-10}y^6 - 2x^2y^{-8} = \frac{8y^{14} - 2x^{12}}{x^{10}y^8}$