Question

if a=5+2√6 and b=1/a, then what will be the value of a^2+b^2 and a^3+b^3

Answer 1

a = 5 +2√6
b = 1/(5+2√6) = (5-2√6)/1 = 5-2√6
a + b = 10
ab = 25 - 24 = 1
(a+b)² = a²+b²+2ab
(10)² = a²+b²+2*1
100 = a²+b²+2
a²+b² = 98
(a+b)³ = a³+b³+3ab(a+b)
(10)³ = a³+b³+3*1*10
1000 = a³+b³+30
a³+b³ = 970

Answer 2

Concept

algebraic identities

(a+b)² = a²+b²+2ab

(a+b)³ = a³+b³+3ab(a+b)

Given

2 variable a and b such that a=5+2√6 and b=1/a

Find

we need to find the value of a^2+b^2 and a^3+b^3

Solution

We have

a = 5 +2√6

b = 1/(5+2√6)

Rationalizing b we get,

(5-2√6)/1 = 5-2√6

Thus,

a + b = 5 +2√6 + 5-2√6

= 10

and ab = (5 +2√6) (5-2√6)

= 25 - 24

= 1

Now,

(a+b)² = a²+b²+2ab

(10)² = a²+b²+ 2*1

100 = a²+b²+ 2

a²+b² = 98

Similarly

(a+b)³ = a³+b³+3ab(a+b)

(10)³ = a³+b³+3*1*10

1000 = a³+b³+30

a³+b³ = 970

Thus, a²+b² = 98 and a³+b³ = 970

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