Đáp án: (Mình đổi đề: $(x-y)^2⇒(x-y)^3;xy+1⇒xy-1$ nhé)
$A=-28$
Giải thích các bước giải:
Ta có:
$x=\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}$
$⇒x^3=(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}})^3$
$=3+2\sqrt{2}+3-2\sqrt{2}+3.\sqrt[3]{3+2\sqrt{2}}.\sqrt[3]{3-2\sqrt{2}}.(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}})$
$=6+3x.\sqrt[3]{(3+2\sqrt{2})(3-2\sqrt{2})}$
$=6+3x.\sqrt[3]{9-8}$
$=6+3x.\sqrt[3]{1}=6+3x$
$⇒x^3-3x=6$
$y=\sqrt[3]{17+12\sqrt{2}}+\sqrt[3]{17-12\sqrt{2}}$
$⇒y^3=(\sqrt[3]{17+12\sqrt{2}}+\sqrt[3]{17-12\sqrt{2}})^3$
$=17+12\sqrt{2}+17-12\sqrt{2}+3.\sqrt[3]{17+12\sqrt{2}}.\sqrt[3]{17-12\sqrt{2}}.(\sqrt[3]{17+12\sqrt{2}}+\sqrt[3]{17-12\sqrt{2}})$
$=34+3y.\sqrt[3]{(17+12\sqrt{2})(17-12\sqrt{2})}$
$=34+3y.\sqrt[3]{289-288}$
$=34+3y.\sqrt[3]{1}=34+3y$
$⇒y^3-3y=34$
Lại có:
$A=(x-y)^3+3(x-y)(xy-1)$
$=(x^3-3x^2y+3xy^2-y^3)+(3x^2y-3x-3xy^2+3y)$
$=x^3-3x^2y+3xy^2-y^3+3x^2y-3x-3xy^2+3y$
$=x^3-3x-y^3+3y$
$=(x^3-3x)-(y^3-3y)$
$=6-34=-28$
