Question

Question for Brainly Teachers​

Answer 1

Step-by-step explanation:

To Solve

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

C1 -> C1-C3

C2 -> C2-C3

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

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

take common (a+b+c) from C1 and C2

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

R3 -> R3-(R1+R2)

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

Take common 2 from R3

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

Expand the determinant

$= 2( {a + b + c)}^{2} ((b + c - a)(c + a - b).ab + a {b}^{2} ) + {a}^{2}(0 + b(c + a - b)) \\ \\ = 2ab( {a + b + c)}^{2}((b + c - a)(c + a) + a(c + a - b)) \\ \\ = 2ab( {a + b + c)}^{2}(bc + ab + {c}^{2} + ac - ac - {a}^{2} + ac + {a}^{2} - ab) \\ \\ = 2ab( {a + b + c)}^{2}(bc + {c}^{2} + ac) \\ \\ = 2abc( {a + b + c)}^{2}(a + b + c) \\ \\ = 2abc {(a + b + c})^{3}$

= RHS

Hence proved

Hope it helps you.