Question

Verify (ab + bc) (ab - bc) + (bs+ ca ) (be - ca ) + (ca + ab) (ca - ab ) = 0​

Answer 1

Answer:

Step-by-step explanation:

(i) \bold {(ab + bc)(ab - bc) + (bc + ca)(bc - ca) + (ca + ab)(ca - ab)}(ab+bc)(ab−bc)+(bc+ca)(bc−ca)+(ca+ab)(ca−ab)

= (ab)^{2} - (bc)^{2} + (bc)^{2} - (ca)^{2} + (ca)^{2} - (ab)^{2}=(ab)2−(bc)2+(bc)2−(ca)2+(ca)2−(ab)2

= (ab)^{2} - (ab)^{2} - (bc)^{2} + (bc)^{2} - (ca)^{2} + (ca)^{2}=(ab)2−(ab)2−(bc)2+(bc)2−(ca)2+(ca)2

= 0=0

Hence, verified.

(ii) \bold {(a+b+c)(a^{2} + b^{2} + c^{2} - ab - bc - ca)}(a+b+c)(a2+b2+c2−ab−bc−ca)

=a(a^{2} + b^{2} + c^{2} - ab - bc - ca)+b(a^{2} + b^{2} + c^{2} - ab - bc - ca)+c(a^{2} + b^{2} + c^{2} - ab - bc - ca)=a(a2+b2+c2−ab−bc−ca)+b(a2+b2+c2−ab−bc−ca)+c(a2+b2+c2−ab−bc−ca)

= a^{3} + ab^{2} + ac^{2} - a^{2}b - abc - a^{2}c + a^{2}b + b^{3} + bc^{2} - ab^{2} - b^{2}c - abc + a^{2}c + b^{2}c + c^{3} - abc - bc^{2} - ac^{2}=a3+ab2+ac2−a2b−abc−a2c+a2b+b3+bc2−ab2−b2c−abc+a2c+b2c+c3−abc−bc2−ac2

= a^{3} + ab^{2} - ab^{2} + ac^{2} - ac^{2} - a^{2}b + a^{2}b - abc - a^{2}c + a^{2}c + b^{3} + bc^{2} - bc^{2} - b^{2}c + b^{2}c - abc + c^{3} - abc=a3+ab2−ab2+ac2−ac2−a2b+a2b−abc−a2c+a2c+b3+bc2−bc2−b2c+b2c−abc+c3−abc

= a^{3}-abc + b^{3} - abc + c^{3} - abc=a3−abc+b3−abc+c3−abc

= a^{3} + b^{3} + c^{3} - abc - abc - abc=a3+b3+c3−abc−abc−abc

= a^{3} + b^{3} + c^{3} - 3abc=a3+b3+c3−3abc

Hence, verified.

(iii) \bold {(p-q)(p^{2} + pq + q^{2})}(p−q)(p2+pq+q2)

=p(p^{2} + pq + q^{2}) -q(p^{2} + pq + q^{2})=p(p2+pq+q2)−q(p