Question

If a+b=1, a2+b2=2 and a3+b3=7, find a5+b5=?​

Answer 1

Given:

a + b = 1, a2 + b2 = 2 and a3 + b3 = 7

To find:

Find a5 + b5 = ?

Solution:

From given, we have,

a + b = 1, a2 + b2 = 2 and a3 + b3 = 7

Use the formula, (a + b)² = a² + b² + 2ab

so, we have, (1²) = 2 + 2ab

1 - 2 = 2ab

-1 = 2 ab

∴ ab = -1/2

(a + b)⁵ = a⁵ + b⁵ + 5a⁴b + 5ab⁴ + 10a³b² + 10a²b³

⇒ (a + b)⁵ = a⁵ + b⁵ + 5ab (a³ + b³) + 10a²b² (a + b)

⇒ (1)⁵ = a⁵ + b⁵ + 5 (-1/2) (7) + 10 (-1/2)² (1)

⇒ 1 = a⁵ + b⁵ - 35/2 + 10/4

⇒ a⁵ + b⁵ = 1 + 35/2 - 10/4

⇒ a⁵ + b⁵ = 1 + 35/2 - 5/2

⇒ a⁵ + b⁵ = 1 + 30/2

⇒ a⁵ + b⁵ = 32/2

∴ a⁵ + b⁵ = 16

Answer 2

Given:

$\bold{a+b=1}\\\\\bold{a^2+b^2=2}\\\\\bold{a^3+b^3}=7$

To find:

$\bold{a^5+b^5 = ?}$

Solution:

let,

$a+b=1.....(i)\\\\a^2+b^2=2.....(ii)\\\\a^3+b^3=7.....(iii)$

Formula:

$\bold{(a+b)^2=a^2+b^2+2ab}\\\\\bold{a^3+b^3=(a+b)(a^2+b^2-ab)}\\\\\bold{a^5+b^5=(a+b)(a^4- a^3b+ a^2b^2- ab^3+ b^4) }$

solve the equation (iii):

$\Rightarrow a^3+b^3=(a+b)(a^2+b^2-ab)\\\\\Rightarrow 7=(1)(2-ab)\\\\\Rightarrow 7=(2-ab)\\\\\Rightarrow 7=2-ab\\\\\Rightarrow ab=-5$

square the equation (ii) and put the value of ab:

$\Rightarrow (a^2+b^2)^2= (a^2)^2+(b^2)^2+2a^2b^2\\\\\Rightarrow (a^2+b^2)^2= (a^4)+(b^4)+2(ab)^2\\\\\Rightarrow (2)^2= (a^4+b^4)+2(-5)^2\\\\\Rightarrow 4= (a^4+b^4)+50\\\\\Rightarrow a^4+b^4=-46$

apply formula:

$\Rightarrow a^5+b^5=(a+b)(a^4- a^3b+ a^2b^2- ab^3+ b^4)\\\\\Rightarrow a^5+b^5=(a+b)(a^4+b^4 +ab(-a^2+ ab- b^2)\\\\\Rightarrow a^5+b^5=(a+b)(a^4+b^4 -ab(a^2+b^2 -ab)\\\\\Rightarrow a^5+b^5=(1)((-46) -(-5)((2) -(-5))\\\\\Rightarrow a^5+b^5=(1)((-46) -(-5)((2+5))\\\\\Rightarrow a^5+b^5=(1)(-41)(7)\\\\\Rightarrow a^5+b^5=-287$

The final value of $\bold{a^5+b^5=-287}$