Question
By solving $(6^0- 7^0) × (6^0+ 7^0)$ we get ________.
✓
Answer
Using law of exponents, $a^0= 1[\because$ a is non-zero integer$]$
$\therefore$ $(6^0- 7^0) × (6^0+ 7^0) = (1 - 1) × (1 + 1)$
$= 0 × 2 = 0$
Hence,
$(6^0- 7^0) × (6^0+ 7^0) = 0$
$\therefore$ $(6^0- 7^0) × (6^0+ 7^0) = (1 - 1) × (1 + 1)$
$= 0 × 2 = 0$
Hence,
$(6^0- 7^0) × (6^0+ 7^0) = 0$
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