Question

ᴩʟᴇᴀꜱᴇ ᴄᴀɴ ꜱᴏᴍᴇᴏɴᴇ ꜱᴏʟᴠᴇ ᴛʜɪꜱ qᴜᴇꜱᴛɪᴏɴ​​

Answer 1

Qᴜᴇsᴛɪᴏɴ :- if a ≠ 0 and (a - 1/a) = 3 ; find :-

• a² + 1/a²
• a³ - 1/a³

Sᴏʟᴜᴛɪᴏɴ :-

→ (a - 1/a) = 3

Squaring Both sides we get,

→ (a - 1/a)² = 3²

Now using (a - b)² = a² + b² - 2ab in LHS,

→ a² + 1/a² - 2 * a * 1/a = 9

→ (a² + 1/a²) - 2 = 9

→ (a² + 1/a²) = 9 + 2

→ (a² + 1/a²) = 11 (Ans.)

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Again,

→ → (a - 1/a) = 3

Cubing Both sides we get,

→ (a - 1/a)³ = 3³

Now using (a - b)³ = a³ - b³ - 3ab(a - b) in LHS,

→ a³ - 1/a³ - 3 * a * 1/a(a - 1/a) = 27

→ a³ - 1/a³ - 3 * (a - 1/a) = 27

Putting value of (a - 1/a) = 3 Now,

→ a³ - 1/a³ - 3 * 3 = 27

→ a³ - 1/a³ - 9 = 27

→ (a³ - 1/a³) = 27 + 9

→ (a³ - 1/a³) = 36 (Ans.)

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Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

$\star{\sf{ \green{ \: \: a - \frac{1}{a} = 3 \: \: \: ,\: where \: a \neq 0 }}}$

${\bf{\blue{\underline{To\:Find:}}}}$

$\star{\sf{ \green{ \: \: {a}^{2} - \frac{1}{ {a}^{2} } =?}}}$

$\star{\sf{ \green{ \: \: {a}^{3} - \frac{1}{ {a}^{3} } =?}}}$

${\bf{\blue{\underline{Now:}}}}$

$: \implies{\sf{ (1). \: {a}- \frac{1}{ {a} }=3 }}$

$:{\sf{ \: Squaring \: Both \: Side }}$

$: \implies{\sf{ \: \bigg( {a} - \frac{1}{ {a} } \bigg) ^{2} = {3}^{2} }}$

$\boxed{\sf{ {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy }}$

$: \implies{\sf{ \: \bigg( {a}^{2}+ \frac{1}{ {a}^{2} } - 2 \times \frac{1}{a} \times a\bigg) = 9}}$

$: \implies{\sf{ \: {a}^{2} + \frac{1}{ {a}^{2} } - 2 \times \frac{1}{ \cancel{a}} \times \cancel{a} = 9}}$

$: \implies{\sf{ \: { {a}^{2} } + \frac{1}{ { {a}^{2} } } = 9 + 2}}$

$: \implies{\sf{ \: { {a}^{2} } + \frac{1}{ { {a}^{2} } } = 11}}$

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$: \implies{\sf{ (2). \: {a}+ \frac{1}{ {a} }=3 }}$

$:{\sf{ \: Cubing \: Both \: Side }}$

$: \implies{\sf{ \: \bigg( {a} - \frac{1}{ {a} } \bigg) ^{3} = {3}^{3} }}$

$\boxed{\sf{ {(x - y)}^{3} = {x}^{3} + {y}^{3} - 3xy(x - y) }}$

$: \implies{\sf{ \: \bigg( {a}^{3}- \frac{1}{ {a}^{3} } - 3 \times \frac{1}{a} \times a(a - \frac{1}{a} )\bigg) = 27}}$

$: \implies{\sf{ \: \bigg( {a}^{3}- \frac{1}{ {a}^{3} } - 3 \times \frac{1}{ \cancel{a}} \times \cancel{a}(3 )\bigg) = 27}}$

$: \implies{\sf{ \: {a}^{3}-\frac{1}{ {a}^{3} } - 3 \times 3 = 27}}$

$: \implies{\sf{ \: {a}^{3}- \frac{1}{ {a}^{3} } - 9 = 27}}$

$: \implies{\sf{ \: {a}^{3}- \frac{1}{ {a}^{3} } = 27+9}}$

$: \implies{\sf{ \: {a}^{3}- \frac{1}{ {a}^{3} } =36}}$