1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# $\left|\begin{array}{ccc}-a\left({b}^{2}+{c}^{2}-{a}^{2}\right)& 2{b}^{3}& 2{c}^{3}\\ 2{a}^{3}& -b\left({c}^{2}+{a}^{2}-{b}^{2}\right)& 2{c}^{3}\\ 2{a}^{3}& 2{b}^{3}& -c\left({a}^{2}+{b}^{2}-{c}^{2}\right)\end{array}\right|=abc{\left({a}^{2}+{b}^{2}+{c}^{2}\right)}^{3}$

Open in App
Solution

## $∆=\left|\begin{array}{ccc}-a\left({b}^{2}+{c}^{2}-{a}^{2}\right)& 2{b}^{3}& 2{c}^{3}\\ 2{a}^{3}& -b\left({c}^{2}+{a}^{2}-{b}^{2}\right)& 2{c}^{3}\\ 2{a}^{3}& 2{b}^{3}& -c\left({a}^{2}+{b}^{2}-{c}^{2}\right)\end{array}\right|\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=abc\left|\begin{array}{ccc}-{b}^{2}-{c}^{2}+{a}^{2}& 2{b}^{2}& 2{c}^{2}\\ 2{a}^{2}& -{c}^{2}-{a}^{2}+{b}^{2}& 2{c}^{2}\\ 2{a}^{2}& 2{b}^{2}& -{a}^{2}-{b}^{2}+{c}^{2}\end{array}\right|\left[\mathrm{Taking}\mathrm{out}a,b\mathrm{and}c\mathrm{common}\mathrm{from}{C}_{1},{C}_{2}\mathrm{and}{C}_{3}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=abc\left|\begin{array}{ccc}{a}^{2}+{b}^{2}+{c}^{2}& 2{b}^{2}& 2{c}^{2}\\ {a}^{2}+{b}^{2}+{c}^{2}& -{c}^{2}-{a}^{2}+{b}^{2}& 2{c}^{2}\\ {a}^{2}+{b}^{2}+{c}^{2}& 2{b}^{2}& -{a}^{2}-{b}^{2}+{c}^{2}\end{array}\right|\left[\mathrm{Applying}{C}_{1}\to {C}_{1}+{C}_{2}+{C}_{3}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=abc\left({a}^{2}+{b}^{2}+{c}^{2}\right)\left|\begin{array}{ccc}1& 2{b}^{2}& 2{c}^{2}\\ 1& -{c}^{2}-{a}^{2}+{b}^{2}& 2{c}^{2}\\ 1& 2{b}^{2}& -{a}^{2}-{b}^{2}+{c}^{2}\end{array}\right|\left[\mathrm{Taking}\mathrm{out}{a}^{2}+{b}^{2}+c\mathrm{common}\mathrm{from}{C}_{1}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=abc\left({a}^{2}+{b}^{2}+{c}^{2}\right)\left|\begin{array}{ccc}1& 2{b}^{2}& 2{c}^{2}\\ 0& -{c}^{2}-{a}^{2}-{b}^{2}& 0\\ 0& 0& -{a}^{2}-{b}^{2}-{c}^{2}\end{array}\right|\left[\mathrm{Applying}{R}_{2}\to {R}_{2}-{R}_{1}\mathrm{and}{R}_{3}\to {R}_{3}-{R}_{1}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=abc\left({a}^{2}+{b}^{2}+{c}^{2}{\right)}^{3}\left[\mathrm{Expanding}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ Hence proved.

Suggest Corrections
1
Join BYJU'S Learning Program
Related Videos
Modulus
MATHEMATICS
Watch in App
Explore more
Join BYJU'S Learning Program