(a)Add: p(p−q),q(q−r) and r(r−p)
(b)Add: 2x(z−x−y) and 2y(z−y−x)
(c)Subtract: 3l(l−4m+5n) from 4l(10n−3m+2l)
(d)Subtract: 3a(a+b+c)−2b(a−b+c) from 4c(−a+b+c)
(a)Add: p(p−q),q(q−r) and r(r−p)
(b)Add: 2x(z−x−y) and 2y(z−y−x)
(c)Subtract: 3l(l−4m+5n) from 4l(10n−3m+2l)
(d)Subtract: 3a(a+b+c)−2b(a−b+c) from 4c(−a+b+c)
(a)First expression =p(p−q)=p2−pq
Second expression =q(q−r)=q2−qr
Third expression =r(r−p)=r2−pr
Adding the three expressions, we obtain
⇒p(p−q)+q(q−r)+r(r−p)=p2−pq+q2−qr+r2−pr
=p2+q2+r2−(pq+qr+rq)
Therefore, the sum of the given expressions is p2+q2+r2−pq−qr−rp.
(b)First expression =2x(z−x−y)=2xz−2x2−2xy
Second expression =2y(z−y−x)=2yz−2y2−2yx
Adding the two expressions, we obtain
⇒2x(z−x−y)+2y(z−y−x)=2xz−2x2−2xy+2yz−2y2−2yx
=2x2−2y2+2xz−2xy−2yx+2yz
=2x2−2y2+2xz−4xy+2yz
Therefore, the sum of the given expressions is 2x2−2y2+2xz−4xy+2yz
(c)First expression =3l(l−4m+5n)=3l2−12lm+15ln
Second expression =4l(10n−3m+2l)=40ln−12lm+8l2
Subtracting these expressions, we obtain
⇒4l(10n−3m+2l)−3l(l−4m+5n)=40ln−12lm+8l2−(3l2−12lm+15ln)
=40ln−12lm+8l2−3l2+12lm−15ln
=8l2−3l2+40ln−15ln−12lm+12lm
=5l22+25ln
Therefore, the result is 5l22+25ln.
(d)First expression =3a(a+b+c)−2b(a−b+c)=3a2+3ab+3ac−2ba+2b2−2bc
=3a2+2b2+ab+3ac−2bc
Second expression 4c(−a+b+c)=−4ac+4bc+4c2
Subtracting these expressions, we obtain,
⇒4c(−a+b+c)−[3a(a+b+c)−2b(a−b+c)]=−4ac+4bc+4c2−[3a2+2b2+ab+3ac−2bc]
=−4ac+4bc+4c2−3a2−2b2−ab−3ac+2bc
=4c2−3a2−2b2−4ac−3ac+2bc+4bc−ab
=4c2−3a2−2b2−7ac+6bc−ab
Therefore, the result is 4c2−3a2−2b2−7ac+6bc−ab.
(a)First expression =p(p−q)=p2−pq
Second expression =q(q−r)=q2−qr
Third expression =r(r−p)=r2−pr
Adding the three expressions, we obtain
⇒p(p−q)+q(q−r)+r(r−p)=p2−pq+q2−qr+r2−pr
=p2+q2+r2−(pq+qr+rq)
Therefore, the sum of the given expressions is p2+q2+r2−pq−qr−rp.
(b)First expression =2x(z−x−y)=2xz−2x2−2xy
Second expression =2y(z−y−x)=2yz−2y2−2yx
Adding the two expressions, we obtain
⇒2x(z−x−y)+2y(z−y−x)=2xz−2x2−2xy+2yz−2y2−2yx
=2x2−2y2+2xz−2xy−2yx+2yz
=2x2−2y2+2xz−4xy+2yz
Therefore, the sum of the given expressions is 2x2−2y2+2xz−4xy+2yz
(c)First expression =3l(l−4m+5n)=3l2−12lm+15ln
Second expression =4l(10n−3m+2l)=40ln−12lm+8l2
Subtracting these expressions, we obtain
⇒4l(10n−3m+2l)−3l(l−4m+5n)=40ln−12lm+8l2−(3l2−12lm+15ln)
=40ln−12lm+8l2−3l2+12lm−15ln
=8l2−3l2+40ln−15ln−12lm+12lm
=5l22+25ln
Therefore, the result is 5l22+25ln.
(d)First expression =3a(a+b+c)−2b(a−b+c)=3a2+3ab+3ac−2ba+2b2−2bc
=3a2+2b2+ab+3ac−2bc
Second expression 4c(−a+b+c)=−4ac+4bc+4c2
Subtracting these expressions, we obtain,
⇒4c(−a+b+c)−[3a(a+b+c)−2b(a−b+c)]=−4ac+4bc+4c2−[3a2+2b2+ab+3ac−2bc]
=−4ac+4bc+4c2−3a2−2b2−ab−3ac+2bc
=4c2−3a2−2b2−4ac−3ac+2bc+4bc−ab
=4c2−3a2−2b2−7ac+6bc−ab
Therefore, the result is 4c2−3a2−2b2−7ac+6bc−ab.
