Question

Please answer these(sums are ticked with pencil) ​

Answer 1

$\large\underline{\sf{Solution-7}}$

$\rm :\longmapsto\:a + \dfrac{1}{a} = p$

and

$\rm :\longmapsto\:a - \dfrac{1}{a} = q$

$\rm :\longmapsto\: {(x + y)}^{2} - {(x - y)}^{2} = 4xy$

$\red{\rm :\longmapsto\:y \: by \: \dfrac{1}{a} \: \: \: \: and \: \: \: \: \: \: x \: by \: a}$

$\rm :\longmapsto\: {\bigg(a + \dfrac{1}{a} \bigg) }^{2} - {\bigg(a - \dfrac{1}{a} \bigg) }^{2} = 4 \times a \times \dfrac{1}{a}$

$\bf :\longmapsto\: {p}^{2} - {q}^{2} = 4$

$\large\underline{\sf{Solution-8}}$

$\rm :\longmapsto\:\dfrac{ {a}^{2} + 1}{a} = 4$

$\rm :\implies\:a + \dfrac{1}{a} = 4$

On cubing both sides, we get

$\rm :\longmapsto\:{\bigg(a + \dfrac{1}{a}\bigg) }^{3} = {4}^{3}$

$\rm :\longmapsto\: {a}^{3} + \dfrac{1}{ {a}^{3} } + 3 \times a \times \dfrac{1}{a}{\bigg(a + \dfrac{1}{a}\bigg) } = 64$

$\rm :\longmapsto\: {a}^{3} + \dfrac{1}{ {a}^{3} } + 3 \times 4 = 64$

$\rm :\longmapsto\: {a}^{3} + \dfrac{1}{ {a}^{3} } + 12 = 64$

$\rm :\longmapsto\: {a}^{3} + \dfrac{1}{ {a}^{3} } = 64 - 12$

$\rm :\longmapsto\: {a}^{3} + \dfrac{1}{ {a}^{3} } = 52$

On multiply by 2, on both sides, we get

$\bf :\longmapsto\: {2a}^{3} + \dfrac{2}{ {a}^{3} } = 104$

$\large\underline{\sf{Solution-9}}$

$\rm :\longmapsto\:x = \dfrac{1}{4 - x}$

$\rm :\longmapsto\:4 - x = \dfrac{1}{ x}$

$\rm :\longmapsto\:4 = \dfrac{1}{ x} + x$

On cubing both sides, we get

$\rm :\longmapsto\:{\bigg(x + \dfrac{1}{x}\bigg) }^{3} = {4}^{3}$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } + 3 \times x \times \dfrac{1}{x}{\bigg(x + \dfrac{1}{x}\bigg) } = 64$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } + 3 \times 4 = 64$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } + 12= 64$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} }= 64 - 12$

$\red{\bf :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} }= 52}$

On squaring both sides, we get

$\rm :\longmapsto\:{\bigg( {x}^{3} + \dfrac{1}{ {x}^{3} } \bigg) }^{2} = {52}^{2}$

$\rm :\longmapsto\: {x}^{6} + \dfrac{1}{ {x}^{6} } + 2 \times {x}^{3} \times \dfrac{1}{ {x}^{3} } = 2704$

$\rm :\longmapsto\: {x}^{6} + \dfrac{1}{ {x}^{6} } + 2 = 2704$

$\rm :\longmapsto\: {x}^{6} + \dfrac{1}{ {x}^{6} } = 2704 - 2$

$\red{\bf :\longmapsto\: {x}^{6} + \dfrac{1}{ {x}^{6} } = 2702}$

$\large\underline{\sf{Solution-10}}$

$\rm :\longmapsto\:x - \dfrac{1}{x} = 3 + 2 \sqrt{2}$

$\rm :\longmapsto\:{\bigg(x - \dfrac{1}{x}\bigg) }^{3} = {(3 + 2 \sqrt{2}) }^{3}$

$\rm :\longmapsto\: {x}^{3} - \dfrac{1}{ {x}^{3} } - 3 \times x \times \dfrac{1}{x}{\bigg(x - \dfrac{1}{x}\bigg) } = 27 + 16 \sqrt{2} + 18 \sqrt{2}(3 + 2 \sqrt{2})$

$\rm :\longmapsto\: {x}^{3} - \dfrac{1}{ {x}^{3} } - 3(3 + 2 \sqrt{2})= 27 + 16 \sqrt{2} + 54 \sqrt{2} + 72$

$\rm :\longmapsto\: {x}^{3} - \dfrac{1}{ {x}^{3} } - 9 - 6 \sqrt{2}= 99 + 70 \sqrt{2}$

$\rm :\longmapsto\: {x}^{3} - \dfrac{1}{ {x}^{3} } = 9 + 6 \sqrt{2} + 99 + 70 \sqrt{2}$

$\rm :\longmapsto\: {x}^{3} - \dfrac{1}{ {x}^{3} } = 108 + 76 \sqrt{2}$

$\rm :\longmapsto\: \dfrac{1}{4} \bigg({x}^{3} - \dfrac{1}{ {x}^{3} } \bigg) = 27 + 19 \sqrt{2}$

$\large\underline{\sf{Solution-11}}$

$\rm :\longmapsto\:x + \dfrac{1}{x} = 3 \dfrac{1}{3}$

$\rm :\longmapsto\:x + \dfrac{1}{x} = \dfrac{10}{3}$

$\rm :\longmapsto\: {\bigg(x + \dfrac{1}{x} \bigg) }^{2} - {\bigg(x - \dfrac{1}{x} \bigg) }^{2} = 4 \times x \times \dfrac{1}{x}$

$\rm :\longmapsto\: {\bigg( \dfrac{10}{3} \bigg) }^{2} - {\bigg(x - \dfrac{1}{x} \bigg) }^{2} = 4$

$\rm :\longmapsto\: {\bigg( \dfrac{10}{3} \bigg) }^{2} - 4 = {\bigg(x - \dfrac{1}{x} \bigg) }^{2}$

$\rm :\longmapsto\: {\bigg( \dfrac{100}{9} \bigg) }- 4 = {\bigg(x - \dfrac{1}{x} \bigg) }^{2}$

$\rm :\longmapsto\: {\bigg( \dfrac{100 - 36}{9} \bigg) } = {\bigg(x - \dfrac{1}{x} \bigg) }^{2}$

$\rm :\longmapsto\: {\bigg( \dfrac{64}{9} \bigg) } = {\bigg(x - \dfrac{1}{x} \bigg) }^{2}$

$\rm :\longmapsto\: {\bigg( \dfrac{8}{3} \bigg) } = {\bigg(x - \dfrac{1}{x} \bigg) }$

On cubing both sides, we get

$\rm :\longmapsto\:{\bigg(x - \dfrac{1}{x}\bigg) }^{3} = {\bigg( \dfrac{8}{3} \bigg) }^{3}$

$\rm :\longmapsto\: {x}^{3} - \dfrac{1}{ {x}^{3} } - 3 \times x \times \dfrac{1}{x}{\bigg(x - \dfrac{1}{x}\bigg) } = \dfrac{512}{27}$

$\rm :\longmapsto\: {x}^{3} - \dfrac{1}{ {x}^{3} } - 3 \times \dfrac{8}{3} = \dfrac{512}{27}$

$\rm :\longmapsto\: {x}^{3} - \dfrac{1}{ {x}^{3} } - 8 = \dfrac{512}{27}$

$\rm :\longmapsto\: {x}^{3} - \dfrac{1}{ {x}^{3} } = \dfrac{512}{27} + 8$

$\rm :\longmapsto\: {x}^{3} - \dfrac{1}{ {x}^{3} } = \dfrac{512 + 216}{27}$

$\red{\rm :\longmapsto\: {x}^{3} - \dfrac{1}{ {x}^{3} } = \dfrac{728}{27}}$

Answer 2

Answer:

Hope it helps you...

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