Question

if a+b=10 and a2 +b2 =58 then the value of a3+b3 is

Answer 1

a+b=10
cubing both the sides
(A+b)3= A3+b3+3ab(a+b)=1000

A2+b2=58

(a+b)=10
sqauring both the sides
A2+b2+2ab=100
58+2ab=100
2ab=42
ab=21

(A+b)3= A3+b3+3ab(a+b)
A3+b3+3(21)(10)=1000
A3+b3=1000-630
A3+b3=370
answer is 370

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Answer 2

Given,

a+b =10, $a^{2} +b^{2} = 58$,

To find,

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

Solution,

Applying the sum of squares formula, that is,

$(a+b)^2 = a^2+ b^2+2ab$, ------(i)

Now,

a+b=10, $a^{2} +b^{2} = 58$,

Substituting these values in (i),

⇒ $10^2 = 58 + 2ab$,

⇒ $2ab = 100-58$,

⇒ $ab = \frac{42}{2}$,

⇒ ab =21.

Now applying another formula,

$(a+b)^3 = a^3+b^3 + 3ab(a+b)$,

Again substituting values of a+b = 10 and ab = 21,

we get,

⇒ $10^3 = a^3+b^3+ 3(21)(10)$,

⇒ $1000 = a^3 + b^3 +630$,

⇒  $a^3+b^3 = 1000-630$,

⇒ $a^3+b^3= 370$.