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a )
`VP= (a+b)^3-3ab(a+b)`
`=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2`
`=a^3+b^3 =VT (đpcm)`
b)
b) Ta có
`VT=a3+b3+c3−3abc`
`=(a+b)3−3ab(a+b)+c3−3abc`
`=[(a+b)3+c3]−3ab(a+b+c)`
`=(a+b+c)[(a+b)2+c2−c(a+b)]−3ab(a+b+c)`
`=(a+b+c)(a2+b2+2ab+c2−ac−bc−3ab)`
`=(a+b+c)(a2+b2+c2−ab−bc−ca)=VP`
a) Ta có:
`VP= (a+b)^3-3ab(a+b)`
`=a^3 + b^3+3ab ( a + b )- 3ab ( a + b )`
`=a^3 + b^3=VT(dpcm)`
b) Ta có
`VT=a^3+b^3+c^3−3abc`
`=(a+b)^3−3ab(a+b)+c^3−3abc`
`=[(a+b)^3+c^3]−3ab(a+b+c)`
`=(a+b+c)[(a+b)^2+c^2−c(a+b)]−3ab(a+b+c)`
`=(a+b+c)(a^2+b^2+2ab+c^2−ac−bc−3ab)`
`=(a+b+c)(a^2+b^2+c^2−ab−bc−ca)=VP`
\(\left(a+b+c\right)^2=a^2+b^2+c^2\)
=>\(a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2\)
=>\(2\left(ab+bc+ac\right)=0\)
=>ab+bc+ac=0
\(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)
=>\(\dfrac{\left(bc\right)^3+\left(ac\right)^3+\left(ab\right)^3}{\left(abc\right)^3}=\dfrac{3}{abc}\)
=>\(\left(bc\right)^3+\left(ac\right)^3+\left(ab\right)^3=3\left(abc\right)^2\)
\(\Leftrightarrow\left(ab+bc\right)^3-3\cdot ab\cdot bc\cdot\left(ab+bc\right)+\left(ac\right)^3=3\left(abc\right)^2\)
=>\(\left(-ac\right)^3-3\cdot ab\cdot bc\cdot\left(-ac\right)+\left(ac\right)^3-3\left(abc\right)^2=0\)
=>\(-a^3c^3+a^3c^3+3a^2b^2c^2-3a^2b^2c^2=0\)
=>0=0(đúng)
\(a,VT=\left(a^2+b^2\right)\left(c^2+d^2\right)=a^2c^2+b^2c^2+a^2d^2+b^2d^2\)
\(VP=\left(ac+bd\right)^2+\left(ad-bc\right)^2=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2=a^2c^2+b^2c^2+a^2d^2+b^2d^2\)
\(\Rightarrow VT=a^2c^2+b^2c^2+a^2d^2+b^2d^2=VP\left(đpcm\right)\)
b, Tham khảo:Chứng minh hằng đẳng thức:(a+b+c)3= a3 + b3 + c3 + 3(a+b)(b+c)(c+a) - Hoc24
a: Ta có: \(a+b+c=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)
Ta có: a+b+c=0
\(\Leftrightarrow\left(a+b+c\right)^3=0\)
\(\Leftrightarrow a^3+b^3+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow a^3+b^3+c^3=3abc\)
b: Ta có: \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow a+b+c=0\)
a) \(a^3+b^3+c^3=3abc\Leftrightarrow\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc=0\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)(đúng do a+b+c = 0)
a: Ta có: a+b+c=0
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)
Ta có: a+b+c=0
\(\Leftrightarrow\left(a+b+c\right)^3=0\)
\(\Leftrightarrow a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow a^3+b^3+c^3=3abc\)
b: Ta có: \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Leftrightarrow a+b+c=0\)
a) Áp dụng nhiều lần công thức \(\left(x+y\right)^3=x^3-y^3+3xy\left(x+y\right)\), ta có:
\(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+c^3+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3-c^3\)
\(=a^3+b^3+3ab\left(a+b\right)+c^3+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=3\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(Đpcm\right)\)
b) Ta có:
\(a^3+b^3+c^3-3abc\)
\(=a^3+3ab\left(a+b\right)+b^2+c^3-3abc-3ab\left(a+b\right)\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc-ab\right)\)
Mình nghĩ bằng thế này mới đúng, bạn chắc ghi sai đề rồi 
a) Ta có: (a + b + c)3 - a3 - b3 - c3 = [ (a + b + c)3 - a3 ] - ( b3 + c3)
= (a + b + c - a) ( a2 + b2 + c2 + 2ab + 2bc + 2ac + a2 + ab + ac + a2) - (b + c) ( b2 - bc + c3)
= (b + c) ( 3a2 + b2 + c2 + 3ab + 2bc + 3ac) - (b + c) ( b2 - bc + c3)
= ( b + c) ( 3a2 + b2 + c2 + 3ab + 2bc + 3ac - b2 + bc - c3)
= ( b + c) ( 3a2 + 3ab + 3bc + 3ac)
= 3 (b + c) [a (a + b) + c (a + b)]
= 3 (b + c) (a + b) (a + c) (đpcm)
\(a)\) Ta có :
\(a+b+c=0\)
\(\Leftrightarrow\)\(\left(a+b+c\right)^3=0^3\)
\(\Leftrightarrow\)\(a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(a+b+c=0\)\(\Rightarrow\)\(\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)
\(\Leftrightarrow\)\(a^3+b^3+c^3+3.\left(-c\right)\left(-a\right)\left(-b\right)=0\)
\(\Leftrightarrow\)\(a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\)\(a^3+b^3+c^3=3abc\) ( đpcm )
Vậy \(a^3+b^3+c^3=3abc\)
Chúc bạn học tốt ~
a, a+b+c=0 => a+b=-c
=>(a+b)3=(-c)3
=>a3+3a2b+3ab2+b3=-c3
=>a3+3ab(a+b)+b3=-c3
Mà a+b=-c
=>a3-3abc+b3=-c3
=>a3+b3+c3=3abc (đpcm)
b, \(P=\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=\frac{a^3}{abc}+\frac{b^3}{abc}+\frac{c^3}{abc}=\frac{a^3+b^3+c^3}{abc}\)
mà a3+b3+c3=3abc (bài a)
\(\Rightarrow P=\frac{3abc}{abc}=3\)
Vậy P=3