if {vmatrix} a+b+2c & a & b \\ c & 2a+b+c & b \\ c & a & a+2b+c {vmatrix} = 2 , then a³+b³+c³ - 3abc =

Correct Answer is : 1 - 3ab - 3bc - 3ca

Correct Answer is : 1 - 3ab - 3bc - 3ca

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AP EAMCET Engineering 21 Apr 2019 Shift 1 Paper

Section: Mathematics

Question : 4 of 160

Marks: +1, -0

\begin{vmatrix} a+b+2c & a & b \\ c & 2a+b+c & b \\ c & a & a+2b+c \end{vmatrix}

a^{3}+b^{3}+c^{3}

- 3abc =

Solution:

Consider the given matrix.

\begin{vmatrix} a+b+2c & a & b \\ c & 2a+b+c & b \\ c & a & a+2b+c \end{vmatrix}=2

Apply the column operation,

C_{1} \rightarrow C_{1}+C_{2}+C_{3}

and take

2(a+b+c)\;\text{ common from }\;C_{1}

2(a-b+c)\begin{vmatrix} 1 & a & b \\ 1 & b+c+2a & b \\ 1 & a & c+a+2b \end{vmatrix}=2

Apply the row operation

R_{3} \rightarrow R_{3}-R_{1}

2(a+b+c)\begin{vmatrix} 1 & a & b \\ 0 & b+c+a & 0 \\ 0 & 0 & c+a+b \end{vmatrix}=2

2(a+b+c)^{3}=2

(a+b+c)^{3}=1

a+b+c=1

Thus,

a^{3}+b^{3}+c^{3}-3abc \;\; = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)

\;\; =1 \cdot (a^{2}+b^{2}+c^{2})-2ab-2bc-2ca-ab-bc-ca

\;\; =1-2a-2bc-2ca-ab-bc-ca

\;\; =1-3ab-3bc-3ca

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