Question

The expansion of ${\left(3x^2-2ax+3a^2\right)}^3$ by using binomial theorem is

• A. $37x^6-74a{x}^5+117a^2{x}^4-116a^3{x}^3+117a^4{x}^2-54a^{5}x+27a^6$
• B. $27x^6-54a{x}^5+117a^2{x}^4-116a^3{x}^3+117a^4{x}^2-54a^{5}x+27a^6$
• C. $37x^6+54a{x}^5-117a^2{x}^4+116a^3{x}^3+117a^2{x}^4-54a^2{x}^5+27a^6$
• D. None of the above

Answer 1

Answer: (B)

${\left(3x^2-2ax+3a^2\right)}^3={\left[3\left(x^2+a^2\right)-2ax\right]}^3$
$={}^3{C}_0{\left\{3\left(x^2+a^2\right)\right\}}^3-{}^3{C}_1{\left\{3\left(x^2+a^2\right)\right\}}^{2}2ax$
$+{}^3{C}_{2}3\left(x^2+a^2\right)(2ax{)}^2-{}^3{C}_3(2ax{)}^3$
$=27{\left(x^2+a^2\right)}^3-3\times 9{\left(x^2+a^2\right)}^2\times 2ax$
$+3\times 3\left(x^2+a^2\right)4a^2{x}^2-8a^3{x}^3$
Now, open the expansion of ${\left(x^2+a^2\right)}^3,{\left(x^2+a^2\right)}^2$, we get
$=27[\left[{}^3{C}_0{\left(x^2\right)}^3+{}^3{C}_1{\left(x^2\right)}^2{a}^2+{}^3{C}_2{x}^2{\left(a^2\right)}^2+{}^3{C}_3{\left(a^2\right)}^3\right]$
$-27\left[{}^2{C}_0{\left(x^2\right)}^2+{}^2{C}_1{x}^2{a}^2+{}^2{C}_2{\left(a^2\right)}^2\right]\times 2ax$
$+9\left(x^2+a^2\right)4a^2{x}^2-8a^3{x}^3$
$=27\left[x^6+3x^4{a}^2+3x^2{a}^4+a^6\right]-27\left[x^4+2x^2{a}^2+a^4\right]2ax$
$+36a^2{x}^2\left(x^2+a^2\right)-8a^3{x}^3$
$=27x^6+81x^4{a}^2+81x^2{a}^4+27a^6-54ax\left(x^4+2x^2{a}^2+a^4\right)$
$+36a^2{x}^4+36a^4{x}^2-8a^3{x}^3$
$=27x^6+117x^4{a}^2+117a^4{x}^2+27a^6$
$-54a{x}^5-54a^{5}x-8a^3{x}^3-108a^3{x}^3$
$=27x^6-54a{x}^5+117a^2{x}^4-116a^3{x}^3$
$+117a^4{x}^2-54a^{5}x+27a^6$