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$x^3 + y^3 = (x + y)(x^2 +y^2 -xy)$

$133 = 7(x^2 + y^2 + 2xy -3xy)$

$19 = (x + y)^2 - 3xy$

$19 = 49 - 3xy$

$30 = 3xy$

$10=xy$

$x + y = 7$ $\Rightarrow (x+y)^3 = 7^3$ $\Rightarrow x^3 + 3x^2y +3xy^2 + y^3 = 343$ We know that $x^3 +y^3 = 133$ So substituting $\Rightarrow 3x^2y + 3xy^2 = 210$ Take out the 3 and factorise $\Rightarrow (x + y)xy = 70$ $\Rightarrow xy = 10$

Before working with simultaneous equations and substitution, which is somewhat time-consuming, I take no shame in having entertained the idea of a shortcut (with the assumption that x and y would take natural values): the only cubic numbers below 133 are 1, 8, 27, 64 and 125. I then noticed that 125 + 8 = 133. Therefore because x^3 and y^3 are 8 and 125 (in any order), xy = 2 * 5 = 10.

$(x+y)^3 = x^3 + y^3 + 3xy(x+y)$ and given that (x+y) = 7 so, $(7)^3 = 133 + 3xy(7)$
$210 = 21xy$ $xy = 10$

We know,

(x+y)^3 = x^3 + y^3 + 3xy(x+y)

or, 7^3 = 133 + 3xy * 7

or, 343-133 = 21xy

or, 210 = 21xy

or, xy = 10

$( x + y ) ^ 3 = 7 ^ 3$

$\Rightarrow (x^2+2xy+y^2)(x+y) = 343$

$\Rightarrow x^3+2x^2y+y^2x + x^2y+2xy^2+y^3 = 343$

$\Rightarrow x^3+y^3 + xy(2x+y+x+2y) = 343$

$\Rightarrow (x^3+y^3) + (xy)(3x+3y) = 343$

$\Rightarrow (x^3+y^3) + (xy)3(x+y) = 343$

$\Rightarrow (133) + (xy)3(7) = 343$

$\Rightarrow xy = \frac {343-133}{(3)(7)}$

$\Rightarrow xy = 10$

$\begin{aligned} (x+y)^3&=x^3+y^3+3xy(x+y)\\\\7^3&=133+3xy(7)\\\\343&=133+21xy\\\\210&=21xy\\\\xy&=10\end{aligned}$

both X, Y <6, as X+Y=7, & x^3 or Y^3 beyond 5 will make X^3+Y^3>133, Now only pair of numbers less than 6 which add up to 7 are 3,4 or 5,2... now only 5,2 pair satisfies the criteria for both the equations

x+y=7 x^3+y^3=133 〖(x+y)〗^3=x^3+3x^2 y+3〖xy〗^2+y^3=7^3=343 3xy(x+y)=343-133=210 7 3 xy=210 21xy=210 xy=10

( $x+y)$ $^3$ = $x$ $^{3}$ + $y$ $^{3}$ + $3xy$ $(x+y)$

So,

$x$ $^{3}$ + $y$ $^ { 3 }$ = $(x+y)$ $^{3}$ - $3xy$ $(x+y)$

$133$ = $7^{3}$ - $3xy(7)$

$133$ = $343$ - $21xy$

$21xy$ = $210$

$xy$ = $10$

X + Y = 7 X³ + Y³ = 133 X³ + Y³ = 125 + 8 X³ + Y³ = 5³ + 2² X³ = 5³ Y³ = 2³ X = 5 and Y = 2

Dude I just tried with 1&6 2&5 Tadaaaa I got it

x+y = 7

(x+y)³ = 7³

x³ + y³ + 3(x+y)(xy) = 343

It admitting also x³+y³ = 133

133 + 3(7)(xy) = 343

3(7)(xy) = 343 - 133

21xy = 210

xy = 10

Used trial and error. So what 2 numbers make 7. 4+3, 5+ 2, 6+1 or 7+0. Tried 4+3, gave me 98 when put into x^3+y^3. Tried 5+2 and gave me 133. Therefore xy=10

$(x+y)^{3}=x^{3}+y^{3}+3x^{2}y+3xy^{2}$

$x^{3}+y^{3}=(x+y)^{3}-(3x^{2}y+3xy^{2})$

$133=7^{3}-(3x^{2}y+3xy^{2})$

$133=343-3xy(x+y)$

$3xy(x+y)=210$

$xy(x+y)=70$

$7xy=70$

$xy=\boxed{10}$

(x+y)^3 = x^3+y^3+3xy(x+y) Hence xy = ([(x+y)^3-(x^3+y^3)]/3xy) Hence xy = 10

For those not as mathematically inclined, there is a simple guess and check. Where x+y=7 and x^3+y^3=133, there are only so many options out there. After enough guessing, we can conclude 5+2=7, and 5^3(which is 125) and 2^3(which is 8) equals 133.

I knew the maths, but it was gonna take too long doing all the calculations. So all I did was possibility and elimination. It's faster:

x + y = 7

This means that there will be only 1 combination numbers for this first equation: 4 + 3 5 + 2 6 + 1

So, if x^3 + y^3 = 133, if I insert the three possibilities into the equation, it should give me 133. So...

(4^3) + (3^3) = 91 ---> Not it

(5^3) + (2^3) = 133 ---> Here we go :)

5x2 = 10

Et voilà

x.x.x+y.y.y=x^2.x+y^2.y=...

As I was careful to indicate in my first post on this site, I'm no expert mathematician. This was how I approached the problem and arrived at the correct answer: Given that x+y = 7, my task was to find two numerals which, when added, would total 7. The sum of the cubes of these same two numerals must also yield 133. My options were: 0+7=7; 1+6=7; 2+5=7; 3+4=7; 4+3=7; 5+2=7; 6+1=7;or 7+0=7. First, I eliminated 0+7 and 1+6 because the sum of their cubes (0 0 0+7 7 7) did not total 133. My next option was 2+5 which, when their cubes were added, gave the following: (2 2 2) + (5 5 5) = 8+125 = 133. This satisfied formulae 1 and 2: x+y = 2+5 = 7; and x^3+y^3 = 133. Finally, x y = 2 5 = 10. ANSWER: 10. NB I did not find it necessary to explore the remaining five options, because the answer was evident at my third option. I could, of course, have stopped counting options after 3+4, since anything after that is a retrograde repetition of earlier options.

There are some people who don't know exactly how to do some of the formulas So if you are a beginner you can just use system of elimination We all know that there are different numbers that can come together to make 7 like 4+3 or 6+1 or 5+2 and more so if you take the time you can obviously solve it just conduct the formula with each pair of numbers and you will realize 5+2 make the best answer so then just Multiply it by 2 and then you will Get 10 But Sharky Kesa did a great job By the way I am actually thirteen I just set it like this because I didn't know the age limit

x3+y3= {x+y}3 - 3xy(x+y}=7x7x7 - 3xy x 7 =343 - 21xy 133 - 343 = - 21xy -210 = -21xy Then xy = -210/-21 = 10

(x+y)³=7³ - Taking the first equation and ^3 the both sides we have: x³+3x²y+3xy²+y³=343 Now we make:

x³+3x²y+3xy²+ y³= 343

-x³ - y³= -133

0x³+3x²y+3xy²+0y³= 210

so:

3xy(x+y) = 210 but x+y = 7

what lead us to:

xy = 210/ (3*7) xy = 10

x + y = 7

(x + y)³ = 343

(x³ + y³) + 3x²y + 3xy² = 343

133 + 3x²y + 3xy² = 343

3x²y + 3xy² = 210

x²y + xy² = 70

xy (x + y) = 70

7xy = 70

xy = 10

well, i just paired up the nos. that added to 7 and then checked them with the cubes

x^3 + y^3 = (x + y)^3 - 3xy(x + y) = 133

-> 7^3 - 3xy.7 = 133 #plug (x + y) = 7

-> 343 - 21xy = 133

-> - 21xy = 133 - 343 = - 210

-> xy = 10

(x+y)^3 =x^(3 )+3x^2 y+3xy^2+y^3

7^3 =x^(3 )+y^3 +3x^2 y+3xy^2

343 =133+3xy(x+y)

343 =133+3xy(7)

343-133=21xy

210 =21xy

∴xy=10

I think I did it the dumb way and just picked a number hence 5 and 2 and put it into the equation getting 5^3+2^3 which equals 133 lol.

I just manually tried it. first by considering a pair of 4&3 but since cube of 4&3 equals 64 + 27=91 so then I tried for 5&2 i.e. 125+8=133 which proved to be correct

We have the following system,

$x + y = 7$

$x³+y³ = 133$

Elevating in both sides $x+y = 7$ , we have:

$(x + y)³ = 7³$

$x³ + 3xy² + 3x²y + y³ = 7³$

Now The System is:

$x³ + 3xy² + 3x²y + y³ = 343³$ (1)

$x³+y³ = 133$ (2)

Subtracting (1) and (2), we have:

$3xy² + 3x²y = 210$

$3xy(x+y) = 210$

We know that $x + y = 7$ , So:

$3xy * 7 = 210$

$xy = 210/21$

$xy = 10$

(x-y)3 =x3+y3+3xy(x+y)

(x+y)(x^2-xy+y^2)=133

7(x^2-xy+y^2)=133

x^2+2xy+y^2=49

x^2+y^2=49-2xy

7(49-3xy)=133

49-3xy=19

xy=10

(x +y)^3 - 3xy (x+y) = x^3 +y^3 or, 7^3- 3.7xy = 133 or, xy = 343- 133 = 10.

taking x=7-y and substituting in equ 2 we get a quadratic equ solving that in x+y=7 we get the values for x and y as 5 and 2 so ans is 10

x3+y3= (x+y)3-3xy(x+y)

x^3 + y^3 = (x+y) [(x+y)^2 - 3xy] 133 = 7 (7^2 - 3xy) 133 = 7 (49 - 3xy) 133/7 = (49 - 3xy) 19 = 49 - 3xy 3xy = 49 - 19 3xy = 30 xy = 10

x^3 + y^3 = 133 then (x + y) (x^2 + y^2 -xy ) = 133 since x+y = 7 then 7( x^2 + y^2 - xy ) 133 7( (x+y)2 - 3xy ) =133 7( 49 -3xy ) = 133 343 + 7xy = 133 -21xy = -210 xy = 10

x^3 + y^3 + 3 x y (x + y) = (x + y)^3

=> 133 + 3 x y (7) = 7^3

=> 19 + 3 x y = 49

=> x y = 10

x^3+y^3= 133 =(x+y)(x^2-xy+y^2)= 7(x^2-xy+y^2) x^2-xy+y^2= 133/7= 19 //+3xy x^2+2xy+y^2= (x+y)^2=7^2= 49=19+3xy 3xy=30 xy=10

6 6 6* = 256 so a number must be less than six...which means automatically 5...and 5 5 5 = 125 and 2 2 2= 8 and their sum 133 and 5+2 = 7 5*2 =10

(x+y)^{ 3 }=(x^{ 3 }+y^{ 3 })+3x^{ 2 }y+3xy^{ 2 }\ (7)^{ 3 }=(133)+3xy(x+y)\ 343=133+3xy(7)\ 210=21xy\ xy=210/21\ xy=10

x + y (5 + 2 or 2 + 5) = 7. xy = 10

(x+y)^3 = x^3 + y^3 + 3x^2 y + 3x y^2

7^3=133+ 3xy(x+y)

343-133=3xy(7)

xy=210/21

xy=10

given x^3+y^3=133; x+y=7 we know the formula (x+y)^3=x^3+y^3+3xy(x+y) substitute the values 7^3=133+3(7)xy on solving we get 10=xy

We know that (x+y)^3=x^3+y^3. 343=133+21xy. 21xy=210,so xy=10.

x^3+y^3=(x+y)^3-exy(x+y)=133; -21xy=-210;xy=10