Question

7. 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 urgently needed will mark them brainliest!!!

Answer 1

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.

Answer 2

Step-by-step explanation:

Answers:

(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