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Question

Show that:
(i) (3x+7)284x=(3x7)2
(ii) (9p5q)2+180pq=(9p+5q)2
(iii) (43m34n)2+2mn=169m2+916n2
(iv) (4pq+3q)2(4pq3q)2=48pq2
(v) (ab)(a+b)+(bc)(b+c)+(ca)(c+a)=0

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Solution

We know that
(a±b)2=a2+b2±2ab
and,
(ab)(a+b)=a2b2

Now,
(i)
(3x+7)284x
=(3x)2+72+2(3x)(7)84x
=(3x)2+7242x
=(3x7)2


(ii)
(9p5q)2+180pq
=(9p)2+(5q)22(9p)(5q)+180pq
=(9p)2+(5q)290pq+180pq
=(9p)2+(5q)2+90pq
=(9p+5q)2

(iii)
(43m34n)2+2mn

=(43m)2+(34n)22(43m)(34n)+2mn

=(43m)2+(34n)22mn+2mn

=169m2+916n2

(iv)
(4pq+3q)2(4pq3q)2
=(4pq)2+(3q)2+2(4pq)(3q)((4pq)2+(3q)22(4pq)(3q))
=(4pq)2+(3q)2+2(4pq)(3q)(4pq)2(3q)2+2(4pq)(3q)
=24pq2+24pq2
=48pq2

(v)
(ab)(a+b)+(bc)(b+c)+(ca)(c+a)
=a2b2+b2c2+c2a2
=0

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