Question

(3х в кубе+6х-4)-3(х вкубе+х+1)=0

Answer 1

$(3x^{3} + 6x-4) -3(x^{3}+x+1) = 0 3 x^{3} +6x-4-3 x^{3} -3x-3= 0 3x - 7 = 0 3x = 7$
$x = \frac{7}{3}$