Large numbers begone

Just split the numerator using the identity $(a^{3}+b^{3})=(a+b)(a^{2}-ab+b^{2})$

$=\frac{(123^{3}+456^{3})}{(123^{2}-123\times 456+456^{2})}$

$=\frac{(123+456)(123^{2}-123\times 456+456^{2})}{(123^{2}-123\times 456+456^{2})}$

$=(123+456) = \boxed{579}$

This question can be solved by using an algebraic identity i.e. $a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})$ Let $a=123$ and $b=456$ Now, let's solve the equation, $\frac{a^{3}+b^{3}}{a^{2}-ab+b^{2}}$ $=\frac{(a+b)(a^{2}-ab+b^{2})}{a^{2}-ab+b^{2}}$ $=a+b$ Substituting the values of $a$ and $b$ , we get, $a+b$ $=123+456$ $=\boxed{579}$

123^3 + 456^3 = (123 + 456)(123^2−123×456+456^2) <=> The result is 123 + 456 = 579

Using formulae a3+b3/(a2-a.b+b2)

$$ Take a look this simple trick: $$ $(x^3+y^3)=(x+y)(x^2-xy+y^2),$ then $\frac{x^3+y^3}{x^2-xy+y^2}=x+y.$ Now, let $\,x=123\,$ and $\,y=456$ . Therefore $$ $\frac{123^3+456^3}{123^2-(123)(456)+456^2}=123+456=\boxed{579}$ $$ $\text{\# }\mathbb{Q}.\mathbb{E}.\mathbb{D}.\text{\#}$

We can solve this question using algebraic identity:

$(a^{3}+b^{3})=(a+b)(a^{2}-ab+b^{2})$

Let $a = 123$ and $b=456$ . So,

$\frac{(a^{3}+b^{3})}{(a^{2}-ab+b^{2})}=\frac{(123^{3}+456^{3})}{(123^{2}-123 \times 456+456^{2})}$

$= \frac{(123+456)(123^{2}-123 \times 456+456^{2})}{(123^{2}-123 \times 456+456^{2})}$

$=123+456=579$

Thus, the answer is $\boxed{579}$

123^3+456^3/123^2-123 456+456^2=(123+456)(123^2+456^2-123 456)/(123^2+456^2-123*456)=123+456=579

The first step is you know that all of them is seem likely and you must simplify it

First

[[\frac{ $(123+456)^{3}$ -3 $\times$ 123 $\times$ 456(123+456)}{ $(123+456)^{2}$ - 3 $\times$ 123 $\times$ 456 ]]

Second pull out the (123+456)

[[\frac{(123+456) $(123+456)^{2}$ - 3 $\times$ 123 $\times$ 456}{ $(123+456)^{2}$ - 3 $\times$ 123 $\times$ 456}]]

Then you can remove it and this the answer

$\boxed{123+456 = 579}$

123+456=579 { use (x^3+y^3)/x^2-xy+y^2) = x+y

Use the identity 123^3 + 456^3 =(123+456) (123^2-123*456 +456^2) So you are left with 123+456 in the numerator which equals 579.

=(123+456)(123^2-123 456+456^2)/(123^2-123 456+456^2) =123+456 =579

BREAK THE ABOVE QUBE LAW & AVOID THE SAME THING.THEN ANS IS:123+456=579

(a^{3}+b^{3})=(a+b)(a^{2}-ab+b^{2}) is one of the formaulas i.e.
( a^{3}+b^{3} ) / ((a^{2}-ab+b^{2}) = (a+b)

when you put a=123 and b=456 you get the above expression which is equivalent to a+b i.e 123+456 i.e. 579

123^{3}+456^{3}/123^{2}-123 456+456^{2} =(123+456)(123^{2}-123 456+456^{2})/(123^{2}-123*456+456^{2}) =123+456 =579

The formula for $a^{3} + b^{3}$ is $(a + b)(a^{2} - ab + b^{2}$

Apply the same in the numerator,

$\frac{(123 + 456)(123^{2} - 123 \times 456 + 456^{2})}{(123^{2} - 123 \times 456 + 456^{2})}$

$= 123 + 456$

$= 579$

That's the answer!