What is a^x(a^{x+1}+b^{x+2})(a^{x+1}-b^{x+2})b^x ?

The solution to a^x(a^{x+1}+b^{x+2})(a^{x+1}-b^{x+2})b^x is (a^{x+1}+b^{x+2})(a^{x+1}-b^{x+2})(ab)^x

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simplify a^x(a^{x+1}+b^{x+2})(a^{x+1}-b^{x+2})b^x

simplify

a^{x} (a^{x + 1} + b^{x + 2}) (a^{x + 1} - b^{x + 2}) b^{x}

(a^{x + 1} + b^{x + 2}) (a^{x + 1} - b^{x + 2}) (ab)^{x}

Solution steps

a^{x} (a^{x + 1} + b^{x + 2}) (a^{x + 1} - b^{x + 2}) b^{x}

Apply exponent rule:

a^{m} b^{m} = (ab)^{m},

assuming

a \geq 0, b \geq 0

a^{x} b^{x} = (ab)^{x}

= (a^{x + 1} + b^{x + 2}) (a^{x + 1} - b^{x + 2}) (ab)^{x}

(1 - t)^{2}

8 1 6^{4}

(x^{2} + 3)^{2}

(e^{(- \frac{1}{100} \cdot 53)})^{2} (e^{(- \frac{1}{135} \cdot 53)})^{3}

(\frac{9}{16})^{\frac{1}{4}} l e f t (\frac{9}{16})^{- 1} (\frac{9}{16})^{\frac{3}{4}}

What is a^x(a^{x+1}+b^{x+2})(a^{x+1}-b^{x+2})b^x ?
The solution to a^x(a^{x+1}+b^{x+2})(a^{x+1}-b^{x+2})b^x is (a^{x+1}+b^{x+2})(a^{x+1}-b^{x+2})(ab)^x