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If (a+b+c)=15 and (ac+bc+ca)=74, the value of (a2+b2+c2)
If (a+b+c)=15 and (ac+bc+ca)=74, the value of (a2+b2+c2)

Resolve into linear factors `a^2+b^2+c^2-ab-bc-ca` - YouTube
Resolve into linear factors `a^2+b^2+c^2-ab-bc-ca` - YouTube

If a2 + b2 + c2 = 24 and ab + bc + ca = -4, then finda+b+c​ - Brainly.in
If a2 + b2 + c2 = 24 and ab + bc + ca = -4, then finda+b+c​ - Brainly.in

Prove the following identities – |(b^2+c^2,ab,ac)(ba,c^2+a^2,bc)(ca,cb,a^2+b ^2)| = 4a^2b^2c^2 ​ - Sarthaks eConnect | Largest Online Education Community
Prove the following identities – |(b^2+c^2,ab,ac)(ba,c^2+a^2,bc)(ca,cb,a^2+b ^2)| = 4a^2b^2c^2 ​ - Sarthaks eConnect | Largest Online Education Community

If a = 2012, b = 2011, c =2010 then the value of a^2 + b^2 + c^2- ab- bc - ca is? - Quora
If a = 2012, b = 2011, c =2010 then the value of a^2 + b^2 + c^2- ab- bc - ca is? - Quora

Using properties of determinants, prove that |(a,b,c)(a2,b2,c2)(bc,ca,ca)| = (a-b)(b-c)(c-a)(ab+bc+ca) - Sarthaks eConnect | Largest Online Education Community
Using properties of determinants, prove that |(a,b,c)(a2,b2,c2)(bc,ca,ca)| = (a-b)(b-c)(c-a)(ab+bc+ca) - Sarthaks eConnect | Largest Online Education Community

اگر a^2 + b^2 + c^2-ab-bc-ca = 0 ، ثابت کنید که a = b = c
اگر a^2 + b^2 + c^2-ab-bc-ca = 0 ، ثابت کنید که a = b = c

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

Question 7 - Show |-a2 ab ac ba -b2 bc ca cb -c2| = 4a2b2b2
Question 7 - Show |-a2 ab ac ba -b2 bc ca cb -c2| = 4a2b2b2

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

|(a^2,bc,ac+c^2),(a^2+ab,b^2,ac),(ab,b^2+bc,c^2)|=4a^2b^2c^2
|(a^2,bc,ac+c^2),(a^2+ab,b^2,ac),(ab,b^2+bc,c^2)|=4a^2b^2c^2

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
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

Using Properties of Determinants, Prove That: `|(3a, -a+B, -a+C),(-b+A, 3b, -b+C),(-c+A, -c+B, 3c)|`= 3(A + B + C) (Ab + Bc + Ca) - Mathematics | Shaalaa.com
Using Properties of Determinants, Prove That: `|(3a, -a+B, -a+C),(-b+A, 3b, -b+C),(-c+A, -c+B, 3c)|`= 3(A + B + C) (Ab + Bc + Ca) - Mathematics | Shaalaa.com

If a+b+c=12, ab+bc+ac=47, what is the meaning of a^2+b^2+c^2? : r/askmath
If a+b+c=12, ab+bc+ac=47, what is the meaning of a^2+b^2+c^2? : r/askmath

If the roots of the equation (c2–ab)x2–2(a2–bc)x + b2–ac = 0 are equal, prove that either a = 0 or a3+ b3+ - Brainly.in
If the roots of the equation (c2–ab)x2–2(a2–bc)x + b2–ac = 0 are equal, prove that either a = 0 or a3+ b3+ - Brainly.in

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
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

If a +b+c =0 then 1/2(a2/bc + b2/ac+c2/ab) is - fk43n7ll
If a +b+c =0 then 1/2(a2/bc + b2/ac+c2/ab) is - fk43n7ll

If a+b+c=4 and a^(2)+b^(2)+c^(2)+3(ab+bc+ac)=21 where a, b, c in R the
If a+b+c=4 and a^(2)+b^(2)+c^(2)+3(ab+bc+ac)=21 where a, b, c in R the

Find the value of a+b+c, if a2 +b2 +c2 = 45 and ab + bc+ac=2.​ - Brainly.in
Find the value of a+b+c, if a2 +b2 +c2 = 45 and ab + bc+ac=2.​ - Brainly.in

a2+b2+c2−ab−bc−ac=0a=5 Find b2+c2.
a2+b2+c2−ab−bc−ac=0a=5 Find b2+c2.

b - c)^{6} + (c - a)^{6} - 3(b - c)^{2}(c - a)^{2} (a - b)^{2} = 2(a^{2} + b ^{2} + c^{2} - - ca - ab)^{3}.
b - c)^{6} + (c - a)^{6} - 3(b - c)^{2}(c - a)^{2} (a - b)^{2} = 2(a^{2} + b ^{2} + c^{2} - - ca - ab)^{3}.

If (a+b+c)=14 and (a2+b2+c2)=74, the value of (ac+bc+ca).
If (a+b+c)=14 and (a2+b2+c2)=74, the value of (ac+bc+ca).

SOLVED] If ab+bc+ca=0 find 1÷a2-bc + 1÷b2-ca + 1÷c2-ab - Self Study 365
SOLVED] If ab+bc+ca=0 find 1÷a2-bc + 1÷b2-ca + 1÷c2-ab - Self Study 365

(a+b+c)(a^(2)+b^(2)+c^(2)-ab-bc-ac)
(a+b+c)(a^(2)+b^(2)+c^(2)-ab-bc-ac)

If a2+b2+c2=250 and ab+bc+ca=3, then a+b+c.
If a2+b2+c2=250 and ab+bc+ca=3, then a+b+c.