5. Show that:
(i)\( (3x + 7)^2 – 84x = (3x – 7)^2\)

(ii) \((9p – 5q)^2 + 180pq = (9p + 5q)^2\)

(iii) \((\frac { 4 }{ 3 } m – \frac { 3 }{ 4 } n)^2 + 2mn = \frac { 16 }{ 9 } m^2 + \frac { 9 }{ 16 } n^2\)

(iv) \((4pq + 3q)^2 – (4pq – 3q)^2 = 48pq^2\)

(v) \((a – b)(a + b) + (b – c) (b + c) + (c – a) (c + a) = 0\)

(i)Given that: \( (3x + 7)^2 – 84x = (3x – 7)^2\)

To show LHS=RHS

LHS:\( (3x + 7)^2 – 84x\)

\(\Rightarrow (3x)^2+2(3x)(7)+7^2-84x\quad\)[using \((a+b)^2=a^2+2ab+b^2\)]

\(\Rightarrow 9x^2+42x+49-84x\)]

\(\Rightarrow 9x^2+42x-84x+49\)]

\(\Rightarrow 9x^2-42x+49\)

\(\Rightarrow (3x)^2-2(3x)(7)+7^2\)

\(\Rightarrow (3x – 7)^2=RHS\)[∵ \((a-b)^2=a^2-2ab+b^2\)]

Hence LHS=RHS

Hence proved
(ii)Given that: \((9p – 5q)^2 + 180pq = (9p + 5q)^2\)

To show LHS=RHS

LHS:\((9p – 5q)^2 + 180pq\)

\(\Rightarrow (9p)^2-2(9p)(5q)+(5q)^2+ 180pq\quad\)[using \((a-b)^2=a^2-2ab+b^2\)]

\(\Rightarrow 81p^2-90pq+25q^2+180pq\)]

\(\Rightarrow 81x^2-90pq+180pq+25q^2\)]

\(\Rightarrow 81x^2+90pq+25q^2\)

\(\Rightarrow (9p)^2+2(9p)(5q)+(5q)^2\)

\(\Rightarrow (9p + 5q)^2=RHS\)[using \((a+b)^2=a^2+2ab+b^2\)]

Hence LHS=RHS

Hence proved
(iii)Given: \((\frac { 4 }{ 3 } m – \frac { 3 }{ 4 } n)^2 + 2mn = \frac { 16 }{ 9 } m^2 + \frac { 9 }{ 16 } n^2\)

To show LHS=RHS

LHS: \((\frac { 4 }{ 3 } m – \frac { 3 }{ 4 } n)^2 + 2mn \)

\(\Rightarrow (\frac { 4 }{ 3 } m)^2-2(\frac { 4 }{ 3 } m)(\frac { 3 }{ 4 } n)+(\frac { 3 }{ 4 } n)^2+ 2mn \quad \)[using \((a-b)^2=a^2-2ab+b^2\)]

\(\Rightarrow \frac { 16 }{ 9 } m^2-2mn+\frac { 9 }{ 16 }n^2+2mn\)]

\(\Rightarrow \frac { 16 }{ 9 } m^2-2mn+2mn+\frac { 9 }{ 16 }n^2\)]

\(\Rightarrow \frac { 16 }{ 9 } m^2+\frac { 9 }{ 16 }n^2\)=RHS

Hence LHS=RHS

Hence proved
(iv)Given: \((4pq + 3q)^2 – (4pq – 3q)^2 = 48pq^2\)

To show LHS=RHS

LHS:\((4pq + 3q)^2 – (4pq – 3q)^2 \)

\(=[(4pq)^2+2(4qp)(3q)+(3q)^2] - [(4pq)^2-2(4pq)(3q)+(3q)^2]\)

\(=[16p^2q^2+24pq^2+9q^2]-[16p^2q^2-24pq^2+9q^2]\)

\(=16p^2q^2+24pq^2+9q^2-16p^2q^2+24pq^2-9q^2\)

\(=(16p^2q^2-16p^2q^2)+(+24pq^2+24pq^2)+(9q^2-9q^2)\quad\)[Placing same terms together]

\(=48pq^2=RHS\)

Hence, LHS=RHS

Hence proved

(v)Given:\((a – b)(a + b) + (b – c) (b + c) + (c – a) (c + a) = 0\)

To show LHS=RHS

LHS:\((a^2-b^2)+(b^2-c^2)+(c^2-a^2)\quad\)[∵ \((x+y)(x-y)=x^2-y^2\)]

\(=a^2-b^2+b^2-c^2+c^2-a^2\)

\(=0=RHS\)

Hence, LHS=RHS

Hence proved