Question

Question 10

Which of the following is equal to 1?
$\left(a\right)2^0+3^0+4^0\left(b\right)2^0\times 3^0\times 4^0\left(c\right)(3^0-2^0)\times 4^0\left(d\right)(3^0-2^0)\times (3^0+2^0)$

Answer 1

Let us solve all the expressions,
For option (a),
$2^0+3^0+4^0=1+1+1=3$.
For option (b),
$2^0\times 3^0\times 4^0=1\times 1\times 1=1$.

For option (c)
$(3^0-2^0)\times 4^0=(1-1)\times 1=0\times 1=0$.

For option (d)
$(3^0-2^0)\times (3^0+2^0)=(1-1)\times (1+1)=0\times 2=0$.

Hence, option (b) is the answer.