A student claims that $\dfrac{1}{x^b} imes \dfrac{1}{x^c} imes \dfra…

Question

A student claims that $\dfrac{1}{x^b}	imes \dfrac{1}{x^c}	imes \dfrac{1}{x^a}	imes \dfrac{1}{x^b}	imes \dfrac{1}{x^a}=1$ for all nonzero $x$. Which condition must hold for this claim to be true?

Accepted Answer

B. $2a+2b+c=0$

Rewrite the product as $x^{-b}x^{-c}x^{-a}x^{-b}x^{-a}=x^{-(2a+2b+c)}$. This equals $1$ for all nonzero $x$ only when the exponent is $0$. Therefore $-(2a+2b+c)=0$, so $2a+2b+c=0$.

Answer

$a+b+c=0$

Answer

$ab+bc+ca=0$

Answer

$a=b=c$

A student claims that $\dfrac{1}{x^b}\times \dfrac{1}{x^c}\times \dfrac{1}{x^a}\times \dfrac{1}{x^b}\times \dfrac{1}{x^a}=1$ for all nonzero $x$. Which condition must hold for this claim to be true?