Using binomial theorem, write down the expansions of the following: `(1-2x+3x^2)^3`

Explanation

Let’s Consider `1-2 x` and `3 x^{2}` as two separate entities

Now,

``
`\ ^{5} C_{0}(1+2 x)^{5}(3 x)^{0}-\ ^{5} C_{1}(1+2 x)^{4}(3 x^{2})^{1}+\ ^{5} C_{2}(1+2 x)^{3}(3 x^{2})^{2}-\ ^{5} C_{3}(1+2 x)^{2}(3 x^{2})^{3}+\ ^{5} C_{4}(1+2 x)^{1}(3 x^{2})^{4}-\ ^{5} C_{5}(1+2 x)^{0}(3 x^{2})^{5}`

`=(1+2 x)^{5}-5(1+2 x)^{4} times 3 x^{2}+10 times(1+2 x)^{3} times 9 x^{4}-10 times(1+2 x)^{2} times 27 x^{6}+5(1+2 x) times 81 x^{8}-243 x^{10}`

`=\ ^{5} C_{0} times(2 x)^{0}+\ ^{5} C_{1} times(2 x)^{1}+\ ^{5} C_{2} times(2 x)^{2}+\ ^{5} C_{3} times(2 x)^{3}+\ ^{5} C_{4} times(2 x)^{4}+\ ^{5} C_{5} times(2 x)^{5}-`

`15 x^{2}[/{4} C_{0}(2 x)^{0}+\ ^{4} C_{1}(2 x)^{1}+\ ^{4} C_{2}(2 x)^{2}+\ ^{4} C_{3}(2 x)^{3}+\ ^{4} C_{4}(2 x)^{4}]+ `

`90 x^{4}[1+8 x^{3}+6 x+12 x^{2}]-270 x^{6}(1+4 x^{2}+4 x)+405 x^{8}+810 x^{9}-243 x^{10}`

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