Question

please answer with explanation and proper steps

Answer 1

ok this is the solution of your question

Answer 2

take common a along R1, b along R2,and c along R3.
so determinant will be,common term abc
$(0 \: \: \: \: \: \: {b}^{2} \: \: \: \: {c}^{2} ) \: \: \: \: row \: \: 1 \\ ( {a}^{2} \: \: \: \: 0 \: \: \: \: {c}^{2} ) \: \: \: \: \: row2 \\ ( {a}^{2} \: \: \: \: \: {b}^{2} \: \: \: 0) \: \: \: \: \: \: row3$
please take all these in the sign of determinant
R3 ---> R3- R1
R2---> R2-R3
$(0 \: \: \: \: \: {b}^{2} \: \: \: {c}^{2} ) \\ (0 \: \: \: \: 0 \: \: \: \: 2 {c}^{2} ) \\ ( {a}^{2} \: \: \: 0 \: \: - {c}^{2} )$
take common a ^2 from C1,b^2 from C2 and c^2 from C3: Now common term outside of the Det. is
${a}^{3} {b}^{3} {c}^{3}$
now inside Det.
$(0 \: \: \: \: 1 \: \: \: \: 1) \\ (0 \: \: \: \: 0 \: \: \: \: 2) \\ (1 \: \: \: \: 0 \: - 1)$
now expand the determinant along C1
0+0+1(2-0)
=2
already we have a common term,so LHS is equal to
$2 {a}^{3} {b}^{3} {c}^{3}$
hope you understand,very well