Question

please solve the question with explanation.(Explanation compulsory)​

Answer 1

Answer :-

Option- b

Given :-

$a + b + c = 4$

$ab + bc + ca = 0$

To find :-

$\left \bigg | \begin{array}{ccc}a&b&c\\b&c&a\\c&a&b\end{array} \bigg\right |$

SOLUTION:-

As they given this

$\left \bigg | \begin{array}{ccc}a&b&c\\b&c&a\\c&a&b\end{array} \bigg\right |$

Firstly we find determinant for this

》 Expansion of along first row

$a\left \bigg|\begin{array}{cc}c&a\\a&b\\\end{array}\right \bigg| - b\left \bigg|\begin{array}{cc}b&a\\c&b\\\end{array}\right \bigg| + c\left \bigg|\begin{array}{cc}b&c\\c&a\\\end{array}\right \bigg|$

Applying ad-bc in order to solve this determinant

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

》 Simplifying the expression

$abc - a {}^{3} - b {}^{3} + abc + abc - c {}^{3}$

》 Keeping the like terms together

$abc + abc + abc - a {}^{3} - b {}^{3} - c {}^{3}$

$3abc - a {}^{3} - b {}^{3} - c {}^{3}$

$- (a {}^{3} + b {}^{3} + c {}^{3} - 3abc)$

As according to the question we know only

a + b + c value and ab+ bc + ac value

From algebraic identities

We know that ,

a³ + b³ + c³ -3abc = (a+b+c)(a² + b² + c² -ab -bc -ca)

So, by using this we can solve the above but we don't know the value of a² + b² + c²

So, we know the algebraic identity

(a + b+c)² = a² + b² + c² + 2ab+2bc + 2ca

(a+b+c)² = a² + b² + c² +2(ab+bc+ca)

a² + b² + c² = (a+b+c)² -2(ab+bc+ca)

Substituting the values ,

a+b+c = 4

ab+bc+ ca = 0

a² + b² + c² = (4)² -2(0)

a² + b² +c² = 16-0

a² + b² + c² = 16

Now we need to get the value of

$- (a {}^{3} + b {}^{3} + c {}^{3} - 3abc)$

So,

a³ + b³ + c³ -3abc = (a+b+c)(a² + b² + c² -ab -bc -ca)

Substituting the values ,

a³ + b³ +c³ -3abc = (4)[16-(ab+bc+ca)]

a³ + b³ +c³ -3abc = 4[16 -0]

a³ + b³ + c³ -3abc = 64

But we need the value of

-(a³+b³+c³-3abc)

-(64)

-64

So, the required answer is -64 option b

Answer 2

Answer:

option (b) is correct

please mark me brainliest