Question

If $\left|\begin{array}{ccc}a& a^3& a^4-1\\ b& b^3& b^4-1\\ c& c^3& c^4-1\end{array}\right|=0$ and $a,b,c$ are all distinct then $abc(ab+bc+ca)$ is equal to

• A. $a+b+c$
• B. $abc$
• C. $0$
• D. none of these

Answer 1

Answer: (A)

$a+b+c$

Given, $\left|\begin{array}{ccc}a& a^3& a^4-1\\ b& b^3& b^4-1\\ c& c^3& c^4-1\end{array}\right|=0$

Doing Row operation $R_1\to R_1-R_2,R_3\to R_3-R_2$

$\left|\begin{array}{ccc}a-b& a^3-b^3& a^4-b^4\\ b& b^3& b^4-1\\ c-b& c^3-b^3& c^4-b^4\end{array}\right|=0$

$\left|\begin{array}{ccc}a-b& (a-b)(a^2+b^2+ab)& (a^2+b^2)(a+b)(a-b)\\ b& b^3& b^4-1\\ c-b& (c-b)(b^2+c^2+cb)& (c^2+b^2)(c+b)(c-b)\end{array}\right|=0$

Taking out $(a-b)$ and $(c-b)$ from the determinant and Doing Row operations $R_1\to R_1-R_3$

$(a-b)(c-b)\left|\begin{array}{ccc}0& a^2-c^2+b(a-c)& (a^3-c^3)+b^2(a-c)+b(a^2-c^2)\\ b& b^3& b^4-1\\ 1& (b^2+c^2+cb)& (c^2+b^2)(c+b)\end{array}\right|=0$

$(a-b)(c-b)(a-c)\left|\begin{array}{ccc}0& a+c+b& (a^2+c^2+ac)+b^2+b(a+c)\\ b& b^3& b^4-1\\ 1& (b^2+c^2+cb)& (c^2+b^2)(c+b)\end{array}\right|=0$

Since $a,b,c$ are distinct.

$\left|\begin{array}{ccc}0& a+c+b& (a^2+c^2+ac)+b^2+b(a+c)\\ b& b^3& b^4-1\\ 1& (b^2+c^2+cb)& (c^2+b^2)(c+b)\end{array}\right|=0$

Doing Row operation $R_2\to R_2-b{R}_3$

$\left|\begin{array}{ccc}0& a+c+b& (a^2+c^2+ac)+b^2+b(a+c)\\ 0& -b{c}^2-c{b}^2& -b{c}^3-c^2{b}^2-b^{3}c-1\\ 1& (b^2+c^2+cb)& (c^2+b^2)(c+b)\end{array}\right|=0$

Doing Row operation $R_2\to R_2+cb{R}_1$

$\left|\begin{array}{ccc}0& a+c+b& (a^2+c^2+ac)+b^2+b(a+c)\\ 0& abc& abc(a+b+c)-1\\ 1& (b^2+c^2+cb)& (c^2+b^2)(c+b)\end{array}\right|=0$

Expanding Determinant, w eget

$abc(a+b+c{)}^2-(a+b+c)-abc(a^2+b^2+c^2)-abc(ab+bc+ac)=0$

$\Rightarrow abc(ab+bc+ac)=abc\left(\right(a+b+c{)}^2-a^2-b^2-c^2)-(a+b+c)$

$\Rightarrow abc(ab+bc+ca)=abc\left(2\right(ab+bc+ac\left)\right)-(a+b+c)$

$\Rightarrow abc(ab+bc+ac)=a+b+c$

Hence, option A is correct.