Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.4 - Page 61: 57

$27x^3-108x^2+144x-64$

$$(3x-4)^3$$ $$=(3x-4)(3x-4)(3x-4)$$ $$=(3x-4)[(3x\times 3x)+(3x\times (-4))+(-4\times 3x)+(-4\times (-4))]$$ $$=(3x-4)(9x^2-12x-12x+16)$$ $$=(3x-4)(9x^2-24x+16)$$ $$=(3x\times 9x^2)+(3x\times (-24x))+(3x\times 16)+(-4\times 9x^2)+(-4\times (-24x))+(-4\times 16)$$ $$=27x^3-72x^2+48x-36x^2+96x-64$$ $$=27x^3-108x^2+144x-64$$