what is a^3+b^3+c^3
Answers
Answer:
Step-by-step explanation:
There are mainy two forms to write the formula of [math]a^3+b^3+c^3[/math]
First one is
[math]a^3+b^3+c^3=(a+b+c)[a^2+b^2+c^2−ab−bc−ac]+3abc[/math]
Second one is
[math]a^3+b^3+c^3=(a+b+c)\frac{1}{2}[(a−b)^2+(b−c)^2+(c-a)^2]+3abc[/math]
To derive these formulas, Follow these steps-
Let me first remind what is [math](a+b+c)^2[/math]
[math](a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)[/math]
[math](a+b+c)^3=(a+b+c)^2(a+b+c)[/math]
Thus after simplification we get,
[math](a+b+c)^3=a^3+b^3+c^3+a^2(b+c)+b^2(a+c)+c^2(a+b)+2(ab+bc+ac)(a+b+c)[/math]
[math]=a^3+b^3+c^3+3a^2(b+c)+3b^2(a+c)+3c^2(a+b)+6abc[/math]
So,
[math](a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(a+c)[/math]
From the last but one step
[math]a^3+b^3+c^3=(a+b+c)^3−[3a^2(b+c)+3b^2(a+c)+3c^2(a+b)+6abc][/math]
So,
[math]a^3+b^3+c^3−3abc=(a+b+c)^3−[3a^2(b+c)+3b^2(a+c)+3c^2(a+b)+9abc][/math]
split the [math]9abc[/math] among the three terms and now collect [math]ab,bc[/math]and [math]ac[/math] terms:
[math]=(a+b+c)3−[3ab(a+b+c)+3bc(a+b+c)+3ac(a+b+c)][/math]
take [math](a+b+c)[/math] as common outside,
[math]=(a+b+c)[a^2+b^2+c^2+2ab+2bc+2ac−3ab−3bc−3ac][/math]
Thus we get
[math]a^3+b^3+c^3−3abc=(a+b+c)[a^2+b^2+c^2−ab−bc−ac][/math]
which may further be rewritten as
[math]a^3+b^3+c^3−3abc=(a+b+c)\frac{1}{2}[(a−b)^2+(b−c)^2+(c-a)^2][/math]
as
[math](a−b)^2=a^2+b^2−2ab[/math] etc.
Most important application of this identity comes when [math]a+b+c=0.[/math]
Clearly the RHS[math]=0[/math], so
[math]a^3+b^3+c^3=3abc[/math]
Also let me add that the sign of the expression [math]a^3+b^3+c^3−3abc[/math] depends purely on the sign of [math]a+b+c[/math], as the term multiplied along is a perfect square and is always [math]≥0.[/math]
Hope that was basic enough.
Thanks for asking and reading.
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Have a great day.
With Regards,
$neha $ingh
There are mainy two forms to write the formula of a3+b3+c3
First one is
a3+b3+c3=(a+b+c)[a2+b2+c2−ab−bc−ac]+3abc
Second one is
a3+b3+c3=(a+b+c)12[(a−b)2+(b−c)2+(c−a)2]+3abc
To derive these formulas, Follow these steps-
Let me first remind what is (a+b+c)2
(a+b+c)2=a2+b2+c2+2(ab+bc+ac)(a+b+c)2=a2+b2+c2+2(ab+bc+ac)
(a+b+c)3=(a+b+c)2(a+b+c)
Thus after simplification we get,(a+b+c)3=a3+b3+c3+a2(b+c)+b2(a+c)+c2(a+b)+2(ab+bc+ac)(a+b+c)
=a3+b3+c3+3a2(b+c)+3b2(a+c)+3c2(a+b)+6abc
So,
(a+b+c)3=a3+b3+c3+3(a+b)(b+c)(a+c)
From the last but one step
a3+b3+c3=(a+b+c)3−[3a2(b+c)+3b2(a+c)+3c2(a+b)+6abc]
So,
a3+b3+c3−3abc=(a+b+c)3−[3a2(b+c)+3b2(a+c)+3c2(a+b)+9abc]
split the 9abc among the three terms and now collect ab,bcand ac terms:
=(a+b+c)3−[3ab(a+b+c)+3bc(a+b+c)+3ac(a+b+c)]
take (a+b+c) as common outside,
=(a+b+c)[a2+b2+c2+2ab+2bc+2ac−3ab−3bc−3ac]
Thus we get
a3+b3+c3−3abc=(a+b+c)[a2+b2+c2−ab−bc−ac]
which may further be rewritten as
a3+b3+c3−3abc=(a+b+c)12[(a−b)2+(b−c)2+(c−a)2]
as
(a−b)2=a2+b2−2ab etc.
Most important application of this identity comes when a+b+c=0.
Clearly the RHS=0, so
a3+b3+c3=3abc
Also let me add that the sign of the expression a3+b3+c3−3abc depends purely on the sign of a+b+c, as the term multiplied along is a perfect square and is always ≥0.
Hope that was basic enough