Question

Simplify the following :

$\left(7^0+3^0\right)\times \left(8^0-5^0\right)$

Answer 1

Given; $\left(7^0+3^0\right)\times \left(8^0-5^0\right)$.

We know that

$a^0=1$.

Now,

$$\begin{gathered} \left(7^0+3^0\right)\times \left(8^0-5^0\right)=\left(1+1\right)\times \left(1-1\right)\left[\because {\mathrm{a}}^0=1\right] \\ =2\times 0 \\ =0 \end{gathered}$$

Hence, $\left(7^0+3^0\right)\times \left(8^0-5^0\right)=0$.