Question

if a2 + 1/a2 = 18
Find:
i) a-1/a
ii)a3 +1/a3​

Answer 1

Step-by-step explanation:

i)a²+ 1/a² =18

deduct 2 on both sides

a²+1/a² -2 = 18-2

2 = 2× a×1/a

a²+1/a² -2*a*1/a = 16

(a - 1/a)² =4²

a-1/a = 4

ii) a²+1/a² =18

add 2 on both sides

a²+1/a² +2*a*1/a = 18+2

(a+1/a)² = 20

(a+1/a) = 2×root5

a³+1/a³ = (a+1/a)(a²+1/a²-a*1/a)

=2*root5×(18-1)

= 2* root5 × 17

= 34*root5

Answer 2

$\sf Given - \\ \\ \sf \: \: \: {a}^{2} + \frac{1}{ {a}^{2} } = 18$

$\sf \: We \: know \: the \: formula, \\ \\ \boxed{ \underline{ \sf{ { {(a - \frac{1}{a}) }^{2} = {a}^{2} + \frac{1}{ {a}^{2} } - 2}}}}$

$\sf \: {i) \: (a - \frac{1}{a} )}^{2} = 18 - 2 \\ \\ \sf \: a - \frac{1}{a} = \sqrt{16} \\ \\ \sf \: = 4 \: units \: ..(ans)$

$\sf \: We \: know \: the \: formula, \\ \\ \boxed{ \underline{ \sf{ {(a - \frac{1}{a}) }^{3} = {a}^{3} + \frac{1}{ {a}^{3} } - 3(a - \frac{1}{a} ) }}}$

$\sf \: ii) \: {(4)}^{3} = {a}^{3} + \frac{1}{ {a}^{3} } - 3(4) \\ \\ \sf \: 64 \: = {a}^{3} + \frac{1}{ {a}^{3} } - 12 \\ \\ \sf \: 64 + 12 = {a}^{3} + \frac{1}{ {a}^{3} } \\ \\ \sf \implies \: {a}^{3} + \frac{1}{ {a}^{3} } = 76 \: units \: ...(ans)$