Đáp án:
Giải thích các bước giải:
a) Do $x+y=a+b$
$⇒(x+y)^2=(a+b)^2$
$⇒x^2+2xy+y^2=a^2+2ab+b^2$
$⇒2xy=2ab$ (do $x^2+y^2=a^2+b^2$)
$⇒xy=ab$
Ta có: $x^3+y^3$
$=(x^3+3x^2y+3xy^2+y^3)-(3x^2y+3xy^2)$
$=(x+y)^3-3xy(x+y)$
$=(a+b)^3-3ab(a+b)$
$=(a^3+3a^2b+3ab^2+b^3)-(3a^2b+3ab^2)$
$=a^3+b^3$ (đpcm)
b) Do $a^2+b^2+(a+b)^2=c^2+d^2+(c+d)^2(1)$
$⇒a^2+b^2+a^2+2ab+b^2=c^2+d^2+c^2+d^2+2cd$
$⇒2a^2+2b^2+2ab=2c^2+2d^2+2cd$
$⇒a^2+b^2+ab=c^2+d^2+cd$
$⇒(a^2+b^2+ab)^2=(c^2+d^2+cd)^2$
$⇒a^4+b^4+a^2b^2+2a^2b^2+2ab^3+2a^3b=c^4+d^4+c^2d^2+2c^2d^2+2cd^3+2c^3d$
$⇒a^4+b^4+3a^2b^2+2ab^3+2a^3b=c^4+d^4+3c^2d^2+2cd^3+2c^3d$
$⇒2a^4+2b^4+6a^2b^2+4ab^3+4a^3b=2c^4+2d^4+6c^2d^2+4cd^3+4c^3d(2)$
Ta có: $[a^2+b^2+(a+b)^2]^2$
$=a^4+b^4+(a+b)^4+2a^2b^2+2a^2(a+b)^2+2b^2(a+b)^2$
$=a^4+b^4+(a+b)^4+2a^2b^2+2a^2(a^2+b^2+2ab)+2b^2(a^2+b^2+2ab)$
$=a^4+b^4+(a+b)^4+(2a^2b^2+2a^4+2a^2b^2+4a^3b+2a^2b^2+2b^4+4ab^3)$
$=a^4+b^4+(a+b)^4+(6a^2b^2+2a^4+4a^3b+2b^4+4ab^3)(3)$
Lại có: $[c^2+d^2+(c+d)^2]^2$
$=c^4+d^4+(c+d)^4+2c^2d^2+2d^2(c+d)^2+2c^2(c+d)^2$
$=c^4+d^4+(c+d)^4+2c^2d^2+2d^2(c^2+2cd+d^2)+2c^2(c^2+2cd+d^2)$
$=c^4+d^4+(c+d)^4+(2c^2d^2+2c^2d^2+4cd^3+2d^4+2c^4+4c^3d+2c^2d^2)$
$=c^4+d^4+(c+d)^4+(6c^2d^2+4cd^3+2d^4+2c^4+4c^3d)(4)$
Từ $(1);(2);(3);(4)⇒a^4+b^4+(a+b)^4=c^4+d^4+(c+d)^4$ (đpcm)
