Đáp án:
`P=2017`
Giải thích các bước giải:
Áp dụng hằng đẳng thức: `(a+b)^3=a^3+b^3+3ab(a+b)`
$x=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}$
$→x^3=9+4\sqrt{5}+9-4\sqrt{5}+3\sqrt[3]{(9+4\sqrt{5})(9-4\sqrt{5})}(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}})$
$→x^3=18+3\sqrt[3]{81-80}.x$
$→x^3=18+3x$
$y=\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}$
$→y^3=3+2\sqrt{2}+3-2\sqrt{2}+3\sqrt[3]{(3+2\sqrt{2})(3-2\sqrt{2})}(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}})$
$→y^3=6+3\sqrt[3]{9-8}.y$
$→y^3=6+3y$
Ta có:
$P=x^3+y^3-3(x+y)+1993$
$→P=18+3x+6+3y-3x-3y+1993$
$→P=2017$