Question

State True or False

$(x^2+a^2{)}^7=(x^7-21x^5{a}^2+35x^3{a}^4-7x{a}^6{)}^2+(7x^{6}a-35x^4{a}^3+21x^2{a}^5-a^7{)}^2$

• A. True
• B. False

Answer 1

Answer: (A)

The correct option is A True

Given $LHS={(x^2+a^2)}^7$

$\therefore LHS={\left(x^2(1+\frac{a^2}{x^2})\right)}^7$

$\therefore LHS=x^{14}{(1+\frac{a^2}{x^2})}^7$

$\therefore LHS=x^{14}[1+7\frac{a^2}{x^2}+\frac{7\times 6}{2!}{\left(\frac{a^2}{x^2}\right)}^2+\frac{7\times 6\times 5}{3!}{\left(\frac{a^2}{x^2}\right)}^3+\frac{7\times 6\times 5\times 4}{4!}{\left(\frac{a^2}{x^2}\right)}^4+\frac{7\times 6\times 5\times 4\times 3}{5!}{\left(\frac{a^2}{x^2}\right)}^5+\frac{7\times 6\times 5\times 4\times 3\times 2}{6!}{\left(\frac{a^2}{x^2}\right)}^6+\frac{7\times 6\times 5\times 4\times 3\times 2\times 1}{7!}{\left(\frac{a^2}{x^2}\right)}^7]$

$\therefore LHS=x^{14}[1+7\frac{a^2}{x^2}+21\frac{a^4}{x^4}+35\frac{a^6}{x^6}+35\frac{a^8}{x^8}+21\frac{a^{10}}{x^{10}}+7\frac{a^{12}}{x^{12}}+\frac{a^{14}}{x^{14}}]$

$\therefore LHS=x^{14}+7a^2{x}^{12}+21a^4{x}^{10}+35a^6{x}^8+35a^8{x}^6+21a^{10}{x}^4+7a^{12}{x}^2+a^{14}$ (1)

$RHS={(x^7-21x^5{a}^2+35x^3{a}^4-7x{a}^6)}^2+{(7x^{6}a-35x^4{a}^3+21x^2{a}^5-a^7)}^2$

$\therefore RHS={[(x^7-21x^5{a}^2)+(35x^3{a}^4-7x{a}^6)]}^2+{[(7x^{6}a-35x^4{a}^3)+(21x^2{a}^5-a^7)]}^2$

$\therefore RHS={(x^7-21x^5{a}^2)}^2+{(35x^3{a}^4-7x{a}^6)}^2+2(x^7-21x^5{a}^2)(35x^3{a}^4-7x{a}^6)+{(7x^{6}a-35x^4{a}^3)}^2+{(21x^2{a}^5-a^7)}^2+2(7x^{6}a-35x^4{a}^3)(21x^2{a}^5-a^7)$

$\therefore RHS={\left(x^7\right)}^2-2x^7\left(21x^5{a}^2\right)+{\left(21x^5{a}^2\right)}^2+{\left(35x^3{a}^4\right)}^2-2\left(35x^3{a}^4\right)\left(7x{a}^6\right)+{\left(7x{a}^6\right)}^2+2(35x^{10}{a}^4-7x^8{a}^6-735x^8{a}^6+147x^6{a}^8)+{\left(7x^{6}a\right)}^2-2\left(7x^{6}a\right)\left(35x^4{a}^3\right)+{\left(35x^4{a}^3\right)}^2+{\left(21x^2{a}^5\right)}^2-2\left(21x^2{a}^5\right)\left(a^7\right)+{\left(a^7\right)}^2+2(147x^8{a}^6-7x^6{a}^8-735x^6{a}^8+35x^4{a}^{10})$

$\therefore RHS=x^{14}-42x^{12}{a}^2+441x^{10}{a}^4+1225x^6{a}^8-490x^4{a}^{10}+49x^2{a}^{12}+2(35x^{10}{a}^4-742x^8{a}^6+147x^6{a}^8)+49x^{12}{a}^2-490x^{10}{a}^4+1225x^8{a}^6+441x^4{a}^{10}-42x^2{a}^{12}+a^{14}+2(147x^8{a}^6-742x^6{a}^8+35x^4{a}^{10})$

$\therefore RHS=x^{14}-42x^{12}{a}^2+441x^{10}{a}^4+1225x^6{a}^8-490x^4{a}^{10}+49x^2{a}^{12}+70x^{10}{a}^4-1484x^8{a}^6+294x^6{a}^8+49x^{12}{a}^2-490x^{10}{a}^4+1225x^8{a}^6+441x^4{a}^{10}-42x^2{a}^{12}+a^{14}+294x^8{a}^6-1484x^6{a}^8+70x^4{a}^{10}$

$\therefore RHS=x^{14}+7x^{12}{a}^2+21x^{10}{a}^4+35x^6{a}^8+21x^4{a}^{10}+7x^2{a}^{12}+35x^8{a}^6+a^{14}$ (2)

From Equation (1) and (2),

${(x^2+a^2)}^7={(x^7-21x^5{a}^2+35x^3{a}^4-7x{a}^6)}^2+{(7x^{6}a-35x^4{a}^3+21x^2{a}^5-a^7)}^2$