Question

$7 {a}^{3} b - 8 {a}^{2} {b}^{2} + 9a \:: \: and \: \: 1 - 6ab$
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Answer 1

$7 {a}^{3} b - 8 {a}^{2} {b}^{2} + 9a = 0$

On cancelling ${a}^{2}$ & ${b}$ from L.H.S we get ,

$7a - 8b + 9a = 0$

On cancelling a from L.H.S

$7 - 8b + 9 = 0 \\ \\ 7 + 9 = 8b \\ \\ 16 = 8b \\ \\ b = \frac{16}{8} \\ \\ b = 2$