Bài 9: \(a^3-3ab^2=2\)
=>\(\left(a^3-3ab^2\right)^2=2^2=4\)
=>\(a^6-6a^4b^2+9a^2b^4=4\)
\(b^3-3a^2b=-11\)
=>\(\left(b^3-3a^2b\right)^2=\left(-11\right)^2=121\)
=>\(b^6-6a^2b^4+9a^4b^2=121\)
Do đó: \(a^6-6a^4b^2+9a^2b^4+b^6-6a^2b^4+9a^4b^2=121+4=125\)
=>\(a^6+3a^4b^2+3a^2b^4+b^6=125\)
=>\(\left(a^2+b^2\right)^3=125=5^3\)
=>\(a^2+b^2=5\)
Bài 10:
\(a+b+c=2\)
=>\(\left(a+b+c\right)^2=2^2=4\)
=>\(a^2+b^2+c^2+2\left(ab+ac+bc\right)=4\)
=>2+2(ab+ac+bc)=4
=>2(ab+ac+bc)=2
=>ab+ac+bc=1
\(a^2+1=a^2+ab+bc+ac\)
=a(a+b)+c(a+b)
=(a+b)(a+c)
\(b^2+1=b^2+bc+ac+ab\)
=b(b+c)+a(b+c)
=(a+b)(b+c)
\(c^2+1=c^2+ac+ab+bc\)
=c(a+c)+b(a+c)
=(a+c)(b+c)
\(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\)
\(=\left(a+b\right)\left(a+c\right)\left(b+c\right)\left(b+a\right)\left(c+a\right)\left(c+b\right)\)
\(=\left\lbrack\left(a+b\right)\left(b+c\right)\left(a+c\right)\right\rbrack^2\) là bình phương của một biểu thức(ĐPCM)
