Prove that ∣∣
∣
∣∣bc−a2ca−b2ab−c2ca−b2ab−c2bc−a2ab−c2bc−a2ca−b2∣∣
∣
∣∣ is divisible by (a+b+c) and find the quotient.
Prove that ∣∣ ∣ ∣∣bc−a2ca−b2ab−c2ca−b2ab−c2bc−a2ab−c2bc−a2ca−b2∣∣ ∣ ∣∣ is divisible by (a+b+c) and find the quotient.
Let Δ=∣∣
∣
∣∣bc−a2ca−b2ab−c2ca−b2ab−c2bc−a2ab−c2bc−a2ca−b2∣∣
∣
∣∣
=∣∣
∣
∣∣bc−a2−ca+b2ca−b2−ab+c2ab−c2ca−b2−ab+c2ab−c2−bc+a2bc−a2ab−c2−bc+a2bc−a2−ca+b2ca−b2∣∣
∣
∣∣
[∵C1→C1−C2 and C2→C2−C3]
=∣∣
∣
∣∣(b−a)(a+b+c)(c−b)(a+b+c)ab−c2(c−b)(a+b+c)(a−c)(a+b+c)bc−a2(a−c)(a+b+c)(b−a)(a+b+c)ca−b2∣∣
∣
∣∣=(a+b+c)2∣∣
∣
∣∣b−ac−bab−c2c−ba−cbc−a2a−cb−aca−b2∣∣
∣
∣∣
[taking (a+b+c) common from C1 and C2 each]
=(a+b+c)2∣∣
∣
∣∣00ab+bc+ca−(a2+b2+c2)c−ba−cbc−a2a−cb−aca−b2∣∣
∣
∣∣
[∵R1→R1+R2+R3]
Now, expanding along R1,
=(a+b+c)2[ab+bc+ca−(a2+b2+c2)(c−b)(b−a)−(a−c)2]=(a+b+c)2(ab+bc+ca−a2−b2−c2)(cb−ac−b2+ab−a2−c2+2ac)=(a+b+c)2(a2+b2+c2−ab−bc−ca)(a2+b2+c2−ac−ab−bc)=12(a+b+c)[(a+b+c)(a2+b2+c2−ab−bc−ca)][(a−b)2+(b−c)2+(c−a)2]=12(a+b+c)(a3+b3+c3−3abc)[(a−b)2+(b−c)2+(c−a)2]
Hence, given determinant is divisible by (a+b+c) and quotient is
(a3+b3+c3−3abc)[(a−b)2+(c−a)2]
Let Δ=∣∣
∣
∣∣bc−a2ca−b2ab−c2ca−b2ab−c2bc−a2ab−c2bc−a2ca−b2∣∣
∣
∣∣
=∣∣
∣
∣∣bc−a2−ca+b2ca−b2−ab+c2ab−c2ca−b2−ab+c2ab−c2−bc+a2bc−a2ab−c2−bc+a2bc−a2−ca+b2ca−b2∣∣
∣
∣∣
[∵C1→C1−C2 and C2→C2−C3]
=∣∣
∣
∣∣(b−a)(a+b+c)(c−b)(a+b+c)ab−c2(c−b)(a+b+c)(a−c)(a+b+c)bc−a2(a−c)(a+b+c)(b−a)(a+b+c)ca−b2∣∣
∣
∣∣=(a+b+c)2∣∣
∣
∣∣b−ac−bab−c2c−ba−cbc−a2a−cb−aca−b2∣∣
∣
∣∣
[taking (a+b+c) common from C1 and C2 each]
=(a+b+c)2∣∣
∣
∣∣00ab+bc+ca−(a2+b2+c2)c−ba−cbc−a2a−cb−aca−b2∣∣
∣
∣∣
[∵R1→R1+R2+R3]
Now, expanding along R1,
=(a+b+c)2[ab+bc+ca−(a2+b2+c2)(c−b)(b−a)−(a−c)2]=(a+b+c)2(ab+bc+ca−a2−b2−c2)(cb−ac−b2+ab−a2−c2+2ac)=(a+b+c)2(a2+b2+c2−ab−bc−ca)(a2+b2+c2−ac−ab−bc)=12(a+b+c)[(a+b+c)(a2+b2+c2−ab−bc−ca)][(a−b)2+(b−c)2+(c−a)2]=12(a+b+c)(a3+b3+c3−3abc)[(a−b)2+(b−c)2+(c−a)2]
Hence, given determinant is divisible by (a+b+c) and quotient is
(a3+b3+c3−3abc)[(a−b)2+(c−a)2]
