Question

(b-3)(b^2+3b+9)-b(b-4)(b+4)=0 7 клас помогите пж

Answer 1

$\displaystyle\bf\\\Big(b-3\Big)\Big(b^{2} +3b+9\Big)-b\Big(b-4\Big)\Big(b+4\Big)=0\\\\\\b^{3} -3^{3} -b\cdot\Big(b^{2} -16\Big)=0\\\\\\b^{3} -27-b^{3} +16b=0\\\\\\16b-27=0\\\\\\16b=27\\\\\\b=1\frac{11}{16}$