The sum of cubes

Let the two numbers be $a$ and $b$ . We have

$a+b=11\cdots(1)$

$a^2+b^2=99\cdots(2)$

Since, upon factorisation,

$a^3+b^3=(a+b)(a^2+b^2-ab)\cdots(3)$

we only have to find $ab$ .

From $(2)$ ,

$a^2+b^2=(a+b)^2-2ab=99\cdots(4)$

Substituting $(1)$ into $(4)$ , we have

$11^2-2ab=99$

$\implies 121-2ab=99$

$\implies 2ab=121-99=22$

$\implies ab=\frac{22}{2}=11$

Substituting $(1),(2)$ and $(4)$ into $(3)$ , we have

$a^3+b^3=11(99-11)$

$=11\times 88=\boxed{968},$

and we are done.

How can a+b and ab = 11.. i got the answer . But ..... ( what values of a and b will satisfy the condition )

Let a , b

Given a+b = 11 & a^2 + b^2 = 99

from (a+b)^2 = a^2 + b^2 + 2ab we will get ab = 11

now (a+b )(a^2 + b^2) = a^3+ b^3 + ab(a+b)

11 * 99 = a^3+ b^3 + 11 * 11 ==> a^3+ b^3 = 11 (99-11) = 11* 88 = 968

When multiplying $(a+b)$ and $(a^2 + b^2)$ we get: $\begin{aligned} (a+b)(a^2 + b^2) = a^3 + a^2b + ab^2 + b^3 \\ = (a^3 + b^3) + ab(a+b) = 11 \cdot 99 \cdots (1) \end{aligned}$ To find $ab$ , we simply square $(a+b)$ and subtract it from $(a^2 + b^2)$ to get $(a^2 + 2ab + b^2) - (a^2 + b^2) = 2ab = 11^2 - 99 = 22 \Rightarrow ab = 11.$ Now we substitute that back into our original equation (1): $\begin{aligned} (a^3 + b^3) + 11(a+b) = (a^3 + b^3) + 11^2 = 11 \cdot 99 \\ a^3 + b^3 = 11 \cdot (99-11) = \boxed{968} \end{aligned}$

$Given\quad that\\ a+b=11\quad ..........\quad (i)\\ { a }^{ 2 }+{ b }^{ 2 }=99\quad ..........\quad (ii)\\ Step\quad I\\ { a }^{ 2 }+{ b }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }+2ab-2ab\\ \quad \quad \quad \quad \quad ={ (a+b) }^{ 2 }-2ab\\ \quad \quad \quad 99={ (11) }^{ 2 }-2ab\qquad \qquad \because \quad { a }^{ 2 }+{ b }^{ 2 }=99\quad and\quad a+b=11\\ \quad \quad \quad 99=121-2ab\\ =>2ab=121-99\\ =>ab=\frac { 22 }{ 2 } =>\quad ab=\boxed { 11 } \\ Step\quad II\\ \qquad Now\quad Find\\ \qquad { a }^{ 3 }+{ b }^{ 3 }=(a+b)({ a }^{ 2 }-ab+{ b }^{ 2 })\\ =>\qquad \quad \quad =(11)\left[ ({ a }^{ 2 }+{ b }^{ 2 })-ab \right] \qquad \because \quad a+b=11\\ =>\qquad \quad \quad =(11)(99-11)\\ =>\qquad \quad \quad =(11)(88)\\ =>{ a }^{ 3 }+{ b }^{ 3 }=\boxed { 968 } \quad Answer$

let x+y=11, x.^2+y.^2=99; then (x+y).^2=x.^2+y.^2+2xy ; 11 11=99+2xy; xy=11; now (x+y).^3=x.^3+y.^3+3xy(x+y) ; 11 11 11=x.^3+y.^3+3 11 11; x.^3+y.^3=11 11(11-3); x.^3+y.^3=121*8; x.^3+y.^3=968;answer

$a^2 +b^2 =(a+b)^2 -2ab=99$

$2ab=121-99=22$

$ab=11$

$a^3 +b^3 =$

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

$=11•(99-11)=11•88=968$

Let's think of $a$ and $b$ as roots of a quadratic polynomial as following: $x^{2}-11x+q=0$ Now, by using Newton's Theorem, we find q:

$S_{2}-11.S_{1}+q.S_{0}=0$ $99-11.11+q.2=0$ $q=11$ Now we know that a and b are roots of the polynomial:

$x^{2}-11x+11=0$

Again, using Newton's Theorem:

$S_{3}-11.S_{2}+11.S_{1}=0$ $S_{3}-11.99+11.11=0$ $S_{3}=88.11.$ $S_{3}=\boxed{968}.$

(x+y)^3 = x^3 + 3xy (x+y) +y^3 sum of cubes = 11^3 - 33xy = 11^3 - 33 ((x+y)^2 - (x^2 + y^2) )/2 = 11^3 - 33 ( 121 - 99 )/2
= 11^3 - 33 * 11 = 11 * (121-33 )
= 88*11 = 968

Can any body explain me if a+b =11 then how ab = 11

It is given that,
a+b = 11 . . . . . . . (1)
a^2+b^2 =99 . . . . .(2)
Now,we know
(a+b)^2 = a^2+2ab+b^2
11^2 = 99 + 2ab . . . . from (1) and (2)
2ab = 22
ab = 11 . . . . . . . .(3)
Now,we know
(a + b)^3 = a^3 + b^3 + 3ab( a + b )
11^3 = a^3 + b^3 + 3(11)(11) . . . . . . from (1),(2) and (3)
1331 = a^3 + b^3 + 363
a^3 + b^3 = 968

First break a^2+b^2 then use a^3+b^3

By factorisation: $a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})$ Now, upon considering the perfect square identity: $ab=\frac{1}{2}((a+b)^{2}-(a^2+b^2))$ $\implies a^{3}+b^{3}=(a+b)(a^{2}+b^{2}-\frac{1}{2}((a+b)^{2}-(a^{2}+b^{2})))$ Substituting in the given values: $a^{3}+b^{3}=(11)(99-\frac{1}{2}(11^{2}-99))$ $=11(99-11)$ $=11\times88=\boxed{968}$

a+b=11 a^2+b^2=99 Therefore,ab=11 therefore,a^3+b^3=968

a+b = 11, hence (a+b)^2 = 121, thus a^2 + b^2 +2ab = 121. and a^2 + b^2 = 99, thus ab =11. Now a^3 + b^3 = (a+b)^3 - 3ab(a+b). Now solving this equation will give a^3 + b^3 = 968.

(a+b)square = (11)square

(a)square+ (b)square +2(a.b) = 121

2(a.b) = 121-99

(a.b) = 22/2

(a.b) = 11

(a+b)cube = (11)cube

(a)cube + (b)cube + 3(a.b)(a+b) = 1331 (formula)

(a)cube + (b)cube + 3(11)(11) = 1331

(a)cube + (b)cube +363 = 1331

(a)cube + (b)cube = 1331-363

(a)cube + (b)cube =968

a+b = 11 ...... equation 1 a^2+b^2 = 99 ...... equation 2

(a+b)^2 = a^2+2ab+b^2 - from equation 1 & 2 11^2 = 99 + 2ab ab = 11..... equation 3

(a+b)^3 = a^3+3ab^2+3a^2b+b^3 from equation 1 & 3 11^3 = a^3+b^3 + 3(11b+11a)

a^3+b^3 = 11^3 - 33(a+b) = 968

Let a+b=11 equation(1) a^2+b^2=99 (a+b)^2+(-2ab)=99 substitute eq.1 〖11〗^2+(-2ab)=99 -2ab=99-121 -2ab=-22 ab=11 equation(2) Now,(a+b)^3=(a+b)(a+b)(a+b) =〖(a〗^2+2ab+b^2)(a+b) =(a^3+3a^2 b+〖3ab〗^2+b^3 ) =(a+b)^3-(3a^2 b+〖3ab〗^2 ) =(a+b)^3-3ab(a+b) Now,substitute the values of eq.1 and 2 =〖11〗^3-3(11)(11) =968

They give us two equations and we will square, subtract and multiply them together to find the answer.

As x+y is 11, you can square that to find (x+y)^2 is 121. Expand that expression to x^2 + 2xy + y^2 and now subtract x^2 + y^2 = 99, to find 2xy=22 and so xy=11.

If we multiply the two equations given, we see 11x99 = x^3 + xy(x + y) + y^3. So 11x99 = the sum of cubes + 11x11. So then the cubes are 11x88 or 968.