If [math]a+b+c=1 [/math], [math]a^2+b^2+c^2=2[/math], and [math]a^3+b^3+c^3=3[/math], then what is [math]a\times b\times c[/math]? - Quora
![a-b)^3 + (b-c)^3 + (c-a)^3=?` (a)`(a+b+c)(a^2+b^2+c^2-ab-bc-ac)` (b)`3(a-b)( b-c)(c-a)` (c)`( - YouTube a-b)^3 + (b-c)^3 + (c-a)^3=?` (a)`(a+b+c)(a^2+b^2+c^2-ab-bc-ac)` (b)`3(a-b)( b-c)(c-a)` (c)`( - YouTube](https://i.ytimg.com/vi/qZ4f3NzbS6Y/maxresdefault.jpg)
a-b)^3 + (b-c)^3 + (c-a)^3=?` (a)`(a+b+c)(a^2+b^2+c^2-ab-bc-ac)` (b)`3(a-b)( b-c)(c-a)` (c)`( - YouTube
![matrices - Prove that the determinant is $(a-b)(b-c)(c-a)(a^2 + b^2 + c^2 )$ - Mathematics Stack Exchange matrices - Prove that the determinant is $(a-b)(b-c)(c-a)(a^2 + b^2 + c^2 )$ - Mathematics Stack Exchange](https://i.stack.imgur.com/WqPIX.jpg)
matrices - Prove that the determinant is $(a-b)(b-c)(c-a)(a^2 + b^2 + c^2 )$ - Mathematics Stack Exchange
![radicals - Use the Cauchy-Schwarz Inequality to prove that $a^2+b^2+c^2 \ge ab+ac+bc $ for all positive $a,b,c$. - Mathematics Stack Exchange radicals - Use the Cauchy-Schwarz Inequality to prove that $a^2+b^2+c^2 \ge ab+ac+bc $ for all positive $a,b,c$. - Mathematics Stack Exchange](https://i.stack.imgur.com/UVS3U.png)
radicals - Use the Cauchy-Schwarz Inequality to prove that $a^2+b^2+c^2 \ge ab+ac+bc $ for all positive $a,b,c$. - Mathematics Stack Exchange
![i) If a^(2)+b^(2)+c^(2)=20 " and" a+b+c=0, " find " ab+bc+ac. (ii) If a^(2)+ b^(2)+c^(2)=250 " and" ab+bc+ca=3, " find" a+b+c. (iii) If a+b+c=11 and ab+ bc+ca=25, then find the value of a^(3)+b^(3)+c^(3)-3 abc. i) If a^(2)+b^(2)+c^(2)=20 " and" a+b+c=0, " find " ab+bc+ac. (ii) If a^(2)+ b^(2)+c^(2)=250 " and" ab+bc+ca=3, " find" a+b+c. (iii) If a+b+c=11 and ab+ bc+ca=25, then find the value of a^(3)+b^(3)+c^(3)-3 abc.](https://d10lpgp6xz60nq.cloudfront.net/question-thumbnail/en_32538276.png)
i) If a^(2)+b^(2)+c^(2)=20 " and" a+b+c=0, " find " ab+bc+ac. (ii) If a^(2)+ b^(2)+c^(2)=250 " and" ab+bc+ca=3, " find" a+b+c. (iii) If a+b+c=11 and ab+ bc+ca=25, then find the value of a^(3)+b^(3)+c^(3)-3 abc.
Prove the following identities –|(-bc,b^2+bc,c^2+bc)(a^2+ac,-ac,c^2+ac)(a^2+ ab,b^2+ab,-ab)| = (ab + bc + ca)^3 - Sarthaks eConnect | Largest Online Education Community
![Simplify: `(a^2 - (b-c)^2)/((a+c)^2 - b^2) + (b^2 - (a-c)^2)/((a+b)^2 - c^2) + (c^2 - (a-b)^2)... - YouTube Simplify: `(a^2 - (b-c)^2)/((a+c)^2 - b^2) + (b^2 - (a-c)^2)/((a+b)^2 - c^2) + (c^2 - (a-b)^2)... - YouTube](https://i.ytimg.com/vi/_5rvzXmlfRk/maxresdefault.jpg)
Simplify: `(a^2 - (b-c)^2)/((a+c)^2 - b^2) + (b^2 - (a-c)^2)/((a+b)^2 - c^2) + (c^2 - (a-b)^2)... - YouTube
![If (b-c)^2,(c-a)^2,(a-b)^2 are in A.P. then (b-c),(c-a),(a-b) are in? | EduRev CA Foundation Question If (b-c)^2,(c-a)^2,(a-b)^2 are in A.P. then (b-c),(c-a),(a-b) are in? | EduRev CA Foundation Question](https://edurev.gumlet.io/ApplicationImages/Temp/540814_83647dd1-162b-4ba8-b4c7-8d0850068567_lg.png)
If (b-c)^2,(c-a)^2,(a-b)^2 are in A.P. then (b-c),(c-a),(a-b) are in? | EduRev CA Foundation Question
How to prove [math]a^2+b^2+c^2-ab-bc-ca[/math] is non-negative for all values of [math] a, b,[/math] and [math]c - Quora
![matrices - Prove that the determinant is $(a-b)(b-c)(c-a)(a^2 + b^2 + c^2 )$ - Mathematics Stack Exchange matrices - Prove that the determinant is $(a-b)(b-c)(c-a)(a^2 + b^2 + c^2 )$ - Mathematics Stack Exchange](https://i.stack.imgur.com/dPZKQ.jpg)
matrices - Prove that the determinant is $(a-b)(b-c)(c-a)(a^2 + b^2 + c^2 )$ - Mathematics Stack Exchange
![prove that a^2 + b^2 + c^2 -ab -bc - ca is always non negative for all values of a, - Maths - Polynomials - 1213071 | Meritnation.com prove that a^2 + b^2 + c^2 -ab -bc - ca is always non negative for all values of a, - Maths - Polynomials - 1213071 | Meritnation.com](https://s3mn.mnimgs.com/img/shared/userimages/mn_images/image/a1(1523).png)