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Algebra Level 4

$\left( \dfrac{(a^0+b^0)(a^0-b^0)}{a^2-b^2} \right)^{0^{4^{5}}}= \ ?$

Assume $a \ne b$ .

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1 -1 not defined 0

1 solution

Abhishek Sharma
May 12, 2015

$\left( \dfrac{(a^0+b^0)(a^0-b^0)}{a^2-b^2} \right)^{0^{4^{5}}}$ $\left( \dfrac{(a^0+b^0)(1-1)}{a^2-b^2} \right)^{0^{4^{5}}}$ $\left( 0 \right)^{0^{4^{5}}}$ $\left( 0 \right)^{0}$

$\left( 0 \right)^{0}$ is not defined.

damn...i am too careless T.T

Zack Yeung - 6 years ago

@Sandeep Bhardwaj Sir , i think it must be mentioned that $a \neq b$ . If $a=b$ then at first sight only , the expression becomes undefined since $a^2-b^2=0$ .

Nihar Mahajan - 6 years, 1 month ago

You're right, but It will not affect the answer. Still I'm adding this to the statement of the problem. Thanks!

Sandeep Bhardwaj - 6 years, 1 month ago

Sir , how do we prove that 0^0 is not defined ?

Sriram Venkatesan - 6 years, 1 month ago

@Sriram Venkatesan – I think you will find this note interesting.

Arulx Z - 6 years, 1 month ago

lol in the scratch pad the value of 0^0 is 1. :p check it out

Ashwin Upadhyay - 6 years ago

@Calvin Lin ,

@Ashwin Upadhyay Is correct. In your scratchpad 0^0 is 1. People can be misguided in such a way. Please change it in the scratchpad too.

tanveen dhingra - 6 years ago

0^0 has baffled mathematicians for years, if you've ever heard of Numberphile they discuss problems like this, in particular 0^0 in their video "Problems with Zero" I believe.

Ben Nunes - 6 years ago

0^0 is equal to 1.

A Former Brilliant Member - 6 years ago

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