Question

Verify the following: i) (ab + bc) (ab – bc) + (bc + ca) (bc – ca) + (ca + ab) (ca – ab) = 0 ii) (a + b + c) (a2 + b2 + c2 – ab – bc – ca) = a3 + b3+ c3 – 3abc iii) (p – q) (p2 + pq + q2) = p3 – q3

Answer 1

i) (ab+bc)(ab-bc) + (bc+ca)(bc-ca) + (ca+ab)(ca-ab) = 0

(ab)2 - (bc)2 + (bc)2 - (ca)2 + (ca)2 - (ab)2 = 0

ab2 - bc2 + bc2 - ca2+ ca2 - ab2 = 0

all will be cancelled out = 0

Hence , LHS = RHS

ii) (a+b+c) (a2 + b2 + c2 - ab - bc - ca) = a3 + b3 + c3 - 3abc

(a + b + c = 0 , a3 + b3 + c3 = 3abc )

(0) ( a2 + b2 + c2 - ab - bc - ca ) = a3 + b3 + c3 - 3abc

0 = a3 + b3 + c3 -3abc

3abc = a3 + b3 + c3

(This maybe wrong according to question)

iii) (p- q) ( p2 + pq + q2) = p3 - q3

( multiply) p3 + p2q  + pq2 - p2q - pq2 - q3 = p3 - q3

( p2q and -p2q , pq2 - pq2 will be cancelled) p3 - q3 = p3 -q3

Hence , LHS = RHS

I hope this answer helps you!

Answer 2

Answers:

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

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

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

$= 0$

Hence, verified.

(ii) $\bold {(a+b+c)(a^{2} + b^{2} + c^{2} - ab - bc - ca)}$

$=a(a^{2} + b^{2} + c^{2} - ab - bc - ca)+b(a^{2} + b^{2} + c^{2} - ab - bc - ca)+c(a^{2} + b^{2} + c^{2} - ab - bc - ca)$

$= a^{3} + ab^{2} + ac^{2} - a^{2}b - abc - a^{2}c + a^{2}b + b^{3} + bc^{2} - ab^{2} - b^{2}c - abc + a^{2}c + b^{2}c + c^{3} - abc - bc^{2} - ac^{2}$

$= a^{3} + ab^{2} - ab^{2} + ac^{2} - ac^{2} - a^{2}b + a^{2}b - abc - a^{2}c + a^{2}c + b^{3} + bc^{2} - bc^{2} - b^{2}c + b^{2}c - abc + c^{3} - abc$

$= a^{3}-abc + b^{3} - abc + c^{3} - abc$

$= a^{3} + b^{3} + c^{3} - abc - abc - abc$

$= a^{3} + b^{3} + c^{3} - 3abc$

Hence, verified.

(iii) $\bold {(p-q)(p^{2} + pq + q^{2})}$

$=p(p^{2} + pq + q^{2}) -q(p^{2} + pq + q^{2})$

$=p^{3} + p^{2}q + pq^{2} - p^{2}q - pq^{2} - q^{3}$

$=p^{3} + p^{2}q - p^{2}q + pq^{2} - pq^{2} - q^{3}$

$= p^{3} - q^{3}$

Hence, verified.