Question

How do you solve (t+1)3=t3+7?

Answer 1

$(t + 1) {}^{3} = {t}^{3} + 7 \\ {t}^{3} + 3 {t}^{2} + 3t + 1 = {t}^{3} + 7 \\ 3 {t}^{2} + 3t - 6 = 0 \\ {t}^{2} + t - 2 = 0$
solve this...
$(t + \frac{1}{2} ) {}^{2} - \frac{1}{4} - 2 = 0 \\ (t + \frac{1}{2} ) {}^{2} = \frac{9}{4} \\ t + \frac{1}{2} = \frac{3}{2} \: or \: \frac{ - 3}{2} \\ t = 1 \: or \: - 2$