Binomial Theorem — MATHS STD 11 Science — Question

Let y =1 - 2x, then

$(1-2\text{x}+3\text{x}^2)^3=(\text{y}+3\text{x}^2)^3$

The expansion of (x + y)ⁿ has n + 1 terms so the expansion of (y + 3x²)³ has 4 terms. Using binomial theorem to expand, we get

$​​(\text{y}+3\text{x}^2)^3={^3\text{C}}_0\text{y}^3(3\text{x}^2)^0+{^3\text{C}}_1\text{y}^2(3\text{x}^2)^1+{^3\text{C}}_2\text{y}(3\text{x}^2)^2+{^3\text{C}}\text{y}^0(3\text{x}^2)^3$

$=\text{y}^3+3\text{y}^2(3\text{x}^2)+3\text{y}(9\text{x}^2)+(27\text{x}^6)$

Substituting y = 1 - 2x, we get,

$(1-2\text{x}+3\text{x}^2)^3=(1-2\text{x})^3+3(1+4\text{x}^2-4\text{x})(3\text{x}^2)+3(1-2\text{x})(9\text{x}^2)+(27\text{x})^6$

$=1-8\text{x}^3-6\text{x}+12\text{x}^2+9\text{x}^2+36\text{x}^4-36\text{x}^3+27\text{x}^2-54\text{x}^3+27\text{x}^6$

$=1-6\text{x}+21\text{x}^2-44\text{x}^3+63\text{x}^4-54\text{x}^5+27\text{x}^6$