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# 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