Question

(3x-1)^3 binomial cube
PLS FAST!!!!

Answer 1

Answer: (3x-1)³=27x^3-27x^2+9x-1

Explanation:

`(3x-1)^3=\\\\(3x)^3-3(3x)^2(1)+3(3x)(1^2)-1^3=\\\\3^3x^3-3(3^2x^2)+9x-1=\\\\27x^3-3(9x^2)+9x-1=\\\\27x^3-27x^2+9x-1\\\\\\(3x-1)^3=\\\\C^9_3(3x)^3(-1)^0+C^1_3(3x)^2(-1)^1+C^2_3(3x)^1(-1)^2+C^3_3(3x)^0(-1)^3=\\\\(3!)/((3-0)!*0!)(3^3x^3)(1)+(3!)/((3-1)^*1!)(3^2x^2)(-1)+(3!)/((3-2)!*2!)(3x)( 1)+(3!)/((3-3)*3!) (1)(-1) =\\\\ (3!)/(3!*1) 27x^3-(2!*3)/(2!*1!) 9x^2+(2!*3)/(1!*2!) 3x-(3!)/(0!3!) =\\\\27x^3-27x^2+9x-1`

Answer 2

27x^3 - 27x^2 + 9x -1.

Explanation:

(3x-1)^3 can be expanded by the binomial theorem;

(a-b)^n = a^n - na^(n-1)*b +n(n-1)a^(n-2)*b^2 - ….. na*b^(n-1)- b^n

So (3x-1)^3 = (3x)^3 - 3*(3x)^2*1 + 3*(3x)*1^2 - 1^3

= 27x^3 - 27x^2 + 9x -1.