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If a, b, c are non-zero real numbers then $|bc ca ab ca ab bc ab bc ca|$ vanishes then,

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\frac{1}{a}

\frac{1}{b}

\frac{1}{c}

= 0

\frac{1}{a}

−

\frac{1}{b}

−

\frac{1}{c}

= 0

\frac{1}{b}

\frac{1}{c}

−

\frac{1}{a}

= 0

\frac{1}{b}

−

\frac{1}{c}

−

\frac{1}{a}

= 0

answer is A.

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Detailed Solution

Given,

|bc ca ab ca ab bc ab bc ca|

vanishes. So, we can write

|bc ca ab ca ab bc ab bc ca|

= 0
Now we know that

|a_{11} a_{12} a_{13} a_{21} a_{22} a_{23} a_{31} a_{32} a_{33}|

a_{11}

|a_{22} a_{23} a_{32} a_{33}|

−

a_{12}

|a_{21} a_{23} a_{31} a_{33}|

a_{13}

|a_{21} a_{22} a_{31} a_{32}|

Let us expand

|bc ca ab ca ab bc ab bc ca|

= 0 along the first row.
Therefore, we can write

bc

|ab bc bc ca|

−

ca

|ca bc ab ca|

ab |ca ab ab bc|

= 0
Also, we know that

|a b c d|

= ad – bc
By using this information, now we can write
bc [(ab) (ca) − (bc) (bc)] – ca [(ca) (ca) − (ab) (bc)] + ab [(ca) (bc) − (ab) (ab)] = 0
Simplifying the above equation, we get
⇒ bc [a2bc − b2c2] – ca [c2a2 − ab2c] + ab [abc2−a2b2] = 0
⇒ a2b2c2 − b3c3 − c3a3 + a2b2c2 + a2b2c2 − a3b3 = 0
Change the sign of each term and add all equal terms in above equation, we get
a3b3 + b3c3 + c3a3 − 3a2b2c2 = 0
Rewrite the above equation, we get
(ab)3 + (bc)3 + (ca)3 − 3(aa)(bb)(cc) = 0
⇒ (ab)3 + (bc)3 + (ca)3 − 3(ab)(bc)(ca) = 0⋯⋯ (i)
Now we are going to use the factorization of a3 + b3 + c3 − 3abc which is given by
a3 + b3 + c3 − 3abc = (a + b + c) (a2 + b2 + c2 + ab + bc + ca).
Let us find the factorization of the LHS of equation (i).
Therefore, we get
(ab + bc + ca) (a2b2 + b2c2 + c2a2 + ab2c + bc2a + a2bc) = 0⋯⋯ (ii)
From equation (ii), we can write either ab + bc + ca = 0⋯⋯ (iii) or
a2b2 + b2c2 + c2a2 + ab2c + bc2a + a2bc = 0⋯⋯ (iv)
Here a, b, c are non-zero real numbers. Let us divide by abc on both sides of equation (iii).
Therefore, we get