Question

If b² + 1/b² = 1, then value of b³+ 1/b³ is,
(A) O
(B) 6
(C) -4
(D) 4​

Answer 1

Answer

The correct option is :

(A) 0

$\large\underline{\sf \bf{Given : }}$

• b² + 1/b² = 1

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

• The value of b³ + 1/b³

$\bf \large{ \underline{Solution : }}$

We are given,

$\longrightarrow\sf {b}^{2} + \dfrac{1}{ {b}^{2} } = 1 \\ \\ \sf \implies {b}^{2} + \dfrac{1}{ {b}^{2} } + 2 = 1 + 2 \\ \\ \sf \implies {b}^{2} + \dfrac{1}{ {b}^{2} } + 2. b.\dfrac{1}{b} = 3 \\ \\ \sf \implies {( b + \dfrac{1}{b}) }^{2} = 3 \\ \\ \sf\implies b + \dfrac{1}{b} = \sqrt{3}$

Now we have from the identities ,

$\tt a^{3} + b^{3} = (a+b)(a^{2}-ab+b^{2})$

Therefore ,

$\sf \hookrightarrow b^{3} + \dfrac{1}{b^{3}} \\\\ \sf\hookrightarrow \{ b + \dfrac{1}{b}\}\{b^{2} - b.\dfrac{1}{b}+{(\dfrac{1}{b})}^{2} \}\\\\ \sf\hookrightarrow ( b + \dfrac{1}{b})(b^{2} + \dfrac{1}{b^{2}} - 1 ) \\\\ \sf\hookrightarrow \sqrt{3}( 1 - 1) \\\\ \sf \hookrightarrow\sqrt{3}\times 0 \\\\ \sf \hookrightarrow 0$

Therefore , the required value of b³ + 1/b³ is 0