Question

if x=a^1/a- a^-1/a ,then prove that x^3+3x= a-1/a

Answer 1

Solution :-

$\sf \: x = a^{\frac{1}{3}} - a^{\frac{-1}{3}} \\ \\ \bf \: cubing \: both \: sides \: we \: get \: \\ \\ \purple\longmapsto\tt \: {x}^{3} = (a^{\frac{1}{3}} - a^{\frac{-1}{3}})^{3} \\ \\ \bf \: now \: using\: \red{\boxed{\bf(x - y)^{3}={x}^{3}-{y}^{3}-3xy(x-y)}} \\ \\ \purple\longmapsto\tt \: {x}^{3}=\: (a^{\frac{1}{3}})^{3} - (a^{\frac{-1}{3}})^{3} - 3 \times a \times \frac{1}{a}(a^{\frac{1}{3}} - a^{\frac{-1}{3}}) \\ \\\purple\longmapsto\tt \: {x}^{3}=a - \frac{1}{a} - 3(a^{\frac{1}{3}} - a^{\frac{-1}{3}}) \\ \\ \bf \: putting \: value \: of \: x \: now \: \\ \\ \purple\longmapsto\tt \: {x}^{3}=a - \frac{1}{a} - 3x \\ \\ \purple\longmapsto\blue{\boxed{\bf{x}^{3} + 3x=(a - \frac{1}{a})}} \: \: \underline{\boxed{\textbf{\red{P}\blue{r}\green{o}\purple{v}\orange{e}\pink{d}}}}$

$\rule{200}{4}$

$\bold{\boxed{\huge{\boxed{\orange{\small{\boxed{\huge{\red{\bold{\huge\bold{\red{\ddot{\smile}}}}}}}}}}}}}$

Answer 2

__________________________

$\huge\tt{GIVEN:}$

• if x = a½ - a -½

__________________________

$\huge\tt{TO~PROOF:}$

• x³+3x = a - ¹/a

__________________________

$\huge\tt{SOLUTION:}$

x = a½ - a -½

Firstly,

Cubing both of the sides here....

↪x³ = (a⅓ - a -⅓)³

using (x-y)³= x³-y³-3xy(x-y)

↪x³ = (a⅓)³ - (a-⅓) ³ - 3 × a × ¹/a (a⅓ - a-⅓)

↪x³ = a - ¹/a - 3 (a⅓ - a -⅓)

Putting the value of x here,

↪x³ = a - ¹/a - 3x

↪x³+ 3 = (a - ¹/a)

Hence,Proved.

__________________________