Question

2) Si x + y = 8 , xy = 4 , calcule: x3 + x2 + y3 + y2 ayuda plzz

Answer 1

1) x + y = 8

2) xy = 4

usamos 1

x = 8 - y

reemplazamos en 2

(8 - y) y = 4

8y - $Y^{2}$ = 4

- $Y^{2}$ + 8y - 4 = 0

usamos fórmula general para ecuaciones de segundo grado

$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

a = -1

b = 8

c = -4

$\frac{-8\pm \sqrt{8^2-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}$

$\frac{-8-\sqrt{8^2-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}$

$\frac{-8-\sqrt{8^2-4\cdot \:1\cdot \:4}}{-2\cdot \:1}$

$\sqrt{8^2-4\cdot \:1\cdot \:4}$

$\sqrt{64-4\cdot \:1\cdot \:4}$

$\sqrt{48}$

$\frac{-8-\sqrt{48}}{-2\cdot \:1}$

$\frac{8+\sqrt{48}}{2}$

$\mathrm{Factorizar}\:8+4\sqrt{3}:\quad 4\left(2+\sqrt{3}\right)$

$\frac{4\left(2+\sqrt{3}\right)}{2}$

Y = $2\left(2+\sqrt{3}\right)$

reemplazamos en 1

$x\:+\:2\left(2+\sqrt{3}\right)\:=\:8$

$\mathrm{Restar\:}2\left(2+\sqrt{3}\right)\mathrm{\:de\:ambos\:lados}$

$x+2\left(2+\sqrt{3}\right)-2\left(2+\sqrt{3}\right)=8-2\left(2+\sqrt{3}\right)$

Simplificar

$x=4-2\sqrt{3}$

$X^{3} + X^{2} + Y^{3} + Y^{2}$

reemplazamos

$\left(4-2\sqrt{3}\right)^3+\left(4-2\sqrt{3}\right)^2+\left(2\left(2+\sqrt{3}\right)\right)^3+\left(2\left(2+\sqrt{3}\right)\right)^2$

$\left(2\left(2+\sqrt{3}\right)\right)^3$ = $2^3\left(2+\sqrt{3}\right)^3$ = $2^3\left(26+15\sqrt{3}\right)$

$\left(2\left(2+\sqrt{3}\right)\right)^2$ = $2^2\left(2+\sqrt{3}\right)^2$ = $2^2\left(7+4\sqrt{3}\right)$

$\left(4-2\sqrt{3}\right)^3+\left(4-2\sqrt{3}\right)^2+2^3\left(26+15\sqrt{3}\right)+2^2\left(7+4\sqrt{3}\right)$

$\left(4-2\sqrt{3}\right)^3+\left(4-2\sqrt{3}\right)^2+8\left(26+15\sqrt{3}\right)+4\left(7+4\sqrt{3}\right)$

$\left(4-2\sqrt{3}\right)^3$

$\mathrm{Aplicar\:la\:formula\:del\:binomio\:al\:cubo}:\quad \left(a-b\right)^3=a^3-3a^2b+3ab^2-b^3$

$a=4,\:\:b=2\sqrt{3}$

$4^3-3\cdot \:4^2\cdot \:2\sqrt{3}+3\cdot \:4\left(2\sqrt{3}\right)^2-\left(2\sqrt{3}\right)^3$

$\mathrm{Simplificar}\:4^3-3\cdot \:4^2\cdot \:2\sqrt{3}+3\cdot \:4\left(2\sqrt{3}\right)^2-\left(2\sqrt{3}\right)^3 = \quad 208-120\sqrt{3}$

$64-96\sqrt{3}+144-24\sqrt{3}$

$64-120\sqrt{3}+144$

$208-120\sqrt{3}$

$208-120\sqrt{3}+\left(4-2\sqrt{3}\right)^2+8\left(26+15\sqrt{3}\right)+4\left(7+4\sqrt{3}\right)$

$\left(4-2\sqrt{3}\right)^2$

$\mathrm{Aplicar\:la\:formula\:del\:binomio\:al\:cuadrado}:\quad \left(a-b\right)^2=a^2-2ab+b^2$

$a=4,\:\:b=2\sqrt{3}$

$4^2-2\cdot \:4\cdot \:2\sqrt{3}+\left(2\sqrt{3}\right)^2$

$\mathrm{Simplificar}\:4^2-2\cdot \:4\cdot \:2\sqrt{3}+\left(2\sqrt{3}\right)^2:\quad 28-16\sqrt{3}$

$16-16\sqrt{3}+12$

$28-16\sqrt{3}$

$208-120\sqrt{3}+28-16\sqrt{3}+8\left(26+15\sqrt{3}\right)+4\left(7+4\sqrt{3}\right)$

$8\left(26+15\sqrt{3}\right)$  $\mathrm{Poner\:los\:parentesis\:utilizando}:\quad \:a\left(b+c\right)=ab+ac$

$a=8,\:b=26,\:c=15\sqrt{3}$

$8\cdot \:26+8\cdot \:15\sqrt{3}$

$\mathrm{Simplificar}\:8\cdot \:26+8\cdot \:15\sqrt{3}:\quad 208+120\sqrt{3}$

$208+120\sqrt{3}$

$208-120\sqrt{3}+28-16\sqrt{3}+208+120\sqrt{3}+4\left(7+4\sqrt{3}\right)$

$4\left(7+4\sqrt{3}\right)$  $\mathrm{Poner\:los\:parentesis\:utilizando}:\quad \:a\left(b+c\right)=ab+ac$

$a=4,\:b=7,\:c=4\sqrt{3}$

$4\cdot \:7+4\cdot \:4\sqrt{3}$

$\mathrm{Simplificar}\:4\cdot \:7+4\cdot \:4\sqrt{3}:\quad 28+16\sqrt{3}$

$28+16\sqrt{3}$

$208-120\sqrt{3}+28-16\sqrt{3}+208+120\sqrt{3}+28+16\sqrt{3}$

$\mathrm{Simplificar}\:208-120\sqrt{3}+28-16\sqrt{3}+208+120\sqrt{3}+28+16\sqrt{3}:\quad 472$

$208-120\sqrt{3}+28-16\sqrt{3}+208+120\sqrt{3}+28+16\sqrt{3}$

$\mathrm{Sumar\:elementos\:similares:}\:-120\sqrt{3}-16\sqrt{3}+120\sqrt{3}+16\sqrt{3}=0$

quedaría

$208+28+208+28$ = 472