Question

If a=3+2√2 find the value of a3/a6+a4+a2+1

Answer 1

Given :-

• a = 3 + 2√2 .

To Find :-

• a³/( a^6 + a⁴ + a² + 1 ) = ?

Solution :-

→ a = (3 + 2√2)

→ 1/a = 1/(3 + 2√2)

→ 1/a = 1/(3 + 2√2) * [ (3 - 2√2) / (3 - 2√2) ]

→ 1/a = (3 - 2√2) / [ (3² - (2√2)² ]

→ 1/a = (3 - 2√2) /( 9 - 8 )

→ 1/a = (3 - 2√2)

So,

→ a + 1/a = (3 + 2√2) + (3 - 2√2)

→ (a + 1/a) = 6 . ------------------------- Equation (1)

Cubing both Sides of Equation one we get,

→ (a + 1/a)³ = 6³

→ a³ + 1/a³ + 3 * a * 1/a *(a + 1/a) = 216

→ a³ + 1/a³ + 3 * 6 = 216

→ a³ + 1/a³ = 216 - 18

→ (a³ + 1/a³) = 198 -------------------- Equation (2).

_____________________________

Now,

→ a³/( a^6 + a⁴ + a² + 1 )

Taking a³ common From Denominator we get,

→ a³ /a³(a³ + a + 1/a + 1/a³)

→ 1/[(a+1/a) + (a³ + 1/a³) ]

Now, Putting Value of Equation (1) & (2), we get ,

→ 1/[6 + 198 ]

→ 1/204 (Ans).

Answer 2

Refers to the attachments

Knowledge Enhancer

• (a+b)²= a²+b²+2ab
• (a-b)²=a²+b²-2ab
• (a+b)(a-b)=a²-b²
• (a+b)³=a³+b³+3ab(a+b)