Question

If a+b=3,a^2+b^2= 40 and find a^3+b^3

Answer 1

Thanks for asking the question!

ANSWER::

Given:-

a+b=3
a²+b²=40
a³+b³ = ?

Lets find ab first by (a+b)²

(a+b)² = a² + b² + 2ab
(3)² = 40 + 2ab
9-40 = 2ab
ab = -31/2 ........ equation 1

Lets use the formula of (a+b)³

(a+b)³ = a³ + b³ + 3ab(a + b)
(3)³ = a³ + b³ + 3.(-31/2).(3)
27=a³+b³-(279/2) 
a³+b³ = 27 + (279/2)
a³+b³ =(54+279)/2
a³+b³ =333/2 or 166.5

Hope it helps!

Answer 2

$\underline{\bold{Given:-}}$

$a + b = 3 \: \: \: \: \: \: \: \: ......(1)\\ \\ {a}^{2} + {b}^{2} = 40 \: \: \: \: \: \: \: .....(2)$

$\underline{\bold{To\:find:-}}$

${a}^{3} + {b}^{3}$

$\underline{\bold{Solution:-}}$

We know that

$\bold{{(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab }\\ \\ from \: eq \: (1) \: and \: (2) \\ \\ {3}^{2} = 40 + 2ab \\ \\ 9 = 40 + 2ab \\ \\ 2ab = 9 - 40 \\ \\ 2ab = - 31 \\ \\ ab = \frac{ - 31}{2} \: \: \: \: \: \: \: ......(3)$

Now , using the identity

$\bold{{(a + b)}^{3} = {a}^{3} + {b}^{3} + 3ab(a + b)} \\ \\ from \: eq \: (1) \: (2) \: and \: (3) \\ \\ {3}^{3} = {a}^{3} + {b}^{3} + 3 \times ( \frac{ - 31}{2} ) \times 3 \\ \\ {a}^{3} + {b}^{3} = 27 + \frac{31 \times 9}{2} \\ \\ {a}^{3} + {b}^{3} = \frac{27 \times 2 + 31 \times9 }{2} \\ \\ {a}^{3} + {b}^{3} = \frac{54 + 279}{2} \\ \\ {a}^{3} + {b}^{3} = \frac{333}{2}$

$\bold{\boxed{{a}^{3} + {b}^{3} =166.5}}$