Question

given( a^2- 7a+ 1)=0 find a^3+1/a^3​

Answer 1

Given :-

• a² - 7 a + 1 = 0

To find :-

• a³ + 1 / a³

Solution :-

given that

→ a² - 7 a + 1 = 0

→ a² + 1 = 7 a  ......eqn (1)

Now,

Let us firstly find

$\longrightarrow\tt{a+\frac{1}{a} }$

so, Taking LCM

$\longrightarrow\tt{\frac{a^2+1}{a}}$

using eqn (1)

$\longrightarrow\tt{\frac{7a}{a}=7}$

Hence,

$\mathrm{a+\frac{1}{a}=7}$  ........ eqn (2)

Now,

we have to find

$\implies\tt{a^3+\frac{1}{a^3}}$

as we know the identity

x³ + y³ = (x + y)³ - 3 x y(x+y)

$\implies\tt{(a+\frac{1}{a})^3-3(a)(\frac{1}{a})(a+\frac{1}{a}}$

$\implies\tt{(a+\frac{1}{a})^3-3(7)}$

using eqn (2)

$\implies\tt{(7)^3-3=343-3(7)=322}$

Hence,

$\boxed{\boxed{\bf{a^3+\frac{1}{a^3}=322}}}$

Answer 2

322

Step-by-step explanation:

Given:

a²- 7a + 1 = 0

To find:

${a}^{3} + \dfrac{1}{ {a}^{3} }$

Solution:

${a}^{2} - 7a + 1 = 0 \\ \\ {a}^{2} + 1 = 7a \\ \\ [\text{Dividing Both side from (a) I get}] \\ \\ a + \frac{1}{a} = 7 \: \: (1) \\ \\ [\text{Clubbing both side }] \\ \\ {(a + \frac{1}{a} )}^{3} = {(7)}^{3} \\ \\ {a}^{3} + \frac{1}{ {a}^{3} } + 3.a. \frac{1}{a}(a + \frac{1}{a} ) = 343 \\ \\ {a}^{3} + \frac{1}{ {a}^{3} } + 3(7) = 343 \\ \\ {a}^{3} + \frac{1}{ {a}^{3} } = 343 - 21 \\ \\{a}^{3} + \frac{1}{ {a}^{3} } = 322$

The required answer is 322

Formula Used

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