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June 4th, 2019, 03:28 PM  #1 
Senior Member Joined: Mar 2019 From: TTF area Posts: 126 Thanks: 1  Algebra what is the actual answer ?
$$(ab)^2=a^2b^2$$ Which values of a and b show that this is not true? 
June 4th, 2019, 03:59 PM  #2 
Member Joined: Oct 2018 From: USA Posts: 89 Thanks: 61 Math Focus: Algebraic Geometry 
$(ab)^{2} = a^{2} 2ab + b^{2} = a^{2}b^{2}$ Subtract $a^{2}$ both sides $2ab = 2b^{2}$ $a=b$ So just choose any $a$ and $b$ such that $a \neq b$ (edit) and $b \neq 0$ as Romsek mentioned Last edited by Greens; June 4th, 2019 at 04:25 PM. 
June 4th, 2019, 04:15 PM  #3 
Senior Member Joined: Sep 2015 From: USA Posts: 2,549 Thanks: 1399 
$(ab)^2 = a^2  2ab + b^2$ $a^2  2ab + b^2 = a^2  b^2$ $2b^2 = 2ab$ $\text{If $b = 0$ this is true $\forall a$}$ $\text{otherwise $b=a$}$ $\text{So the original equation is only true if $b=0 \vee a=b$}$ 
June 4th, 2019, 10:06 PM  #4 
Senior Member Joined: Mar 2019 From: TTF area Posts: 126 Thanks: 1 
What if: $$(a â€“ b)^2= a^2â€“ b^2$$ Which values of a and b show that this is not true? A. a = 1, b = 0 $\hspace{1cm}$ B. a = 1, b = 1 C. a = â€“1, b = â€“1 $\hspace{0.58cm}$ D. a = â€“1, b = 0 E. a = 1, b = â€“1 Thanks for those replies. How about this? Last edited by skipjack; June 5th, 2019 at 08:04 AM. 
June 5th, 2019, 03:53 AM  #5 
Senior Member Joined: Mar 2019 From: iran Posts: 318 Thanks: 14 
E

June 5th, 2019, 05:28 AM  #6 
Senior Member Joined: Oct 2009 Posts: 863 Thanks: 328 
You seriously can't plug in the values and see whether it equals??


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