![The value of the determinant | b^2 - ab b - c bc - ac | ab - b^2 a - b b^2 - ab | bc - ac c - a ab - b^2 The value of the determinant | b^2 - ab b - c bc - ac | ab - b^2 a - b b^2 - ab | bc - ac c - a ab - b^2](https://d1hj4to4g9ba46.cloudfront.net/questions/2022827_1148955_ans_b6064ae55b644c44a4e9eb584afd2090.jpeg)
The value of the determinant | b^2 - ab b - c bc - ac | ab - b^2 a - b b^2 - ab | bc - ac c - a ab - b^2
![Suppose A, B, C are defined as A = a^(2)b + ab^(2) - a^(2)c - ac^(2), B = b^(2)c + bc^(2) - a^(2)b - ab^(2), and C = a^(2)c + ac^(2) - Suppose A, B, C are defined as A = a^(2)b + ab^(2) - a^(2)c - ac^(2), B = b^(2)c + bc^(2) - a^(2)b - ab^(2), and C = a^(2)c + ac^(2) -](https://d10lpgp6xz60nq.cloudfront.net/web-thumb/644180564_web.png)
Suppose A, B, C are defined as A = a^(2)b + ab^(2) - a^(2)c - ac^(2), B = b^(2)c + bc^(2) - a^(2)b - ab^(2), and C = a^(2)c + ac^(2) -
Using properties of determinants, prove that |(-bc,b^2+bc,c^2+bc)(a^2+ac,-ac ,c^2+ac)(a^2+ab,b^2+ab,-ab)| = (ab+bc+ac)^3. - Sarthaks eConnect | Largest Online Education Community
![The determinant |{:(b^2-ab, b-c, bc-ac), (a b-a^2, a-b, b^2-ab) ,(b c-c a, c-a, a b-a^2):}| equals (a)a b c\ (b-c)(c-a)(a-b) (b) (b-c)(c-a)(a-b) (c) (a+b+c)(b-c)(c-a)(a-b) (d) none of these The determinant |{:(b^2-ab, b-c, bc-ac), (a b-a^2, a-b, b^2-ab) ,(b c-c a, c-a, a b-a^2):}| equals (a)a b c\ (b-c)(c-a)(a-b) (b) (b-c)(c-a)(a-b) (c) (a+b+c)(b-c)(c-a)(a-b) (d) none of these](https://d10lpgp6xz60nq.cloudfront.net/web-thumb/642579426_web.png)
The determinant |{:(b^2-ab, b-c, bc-ac), (a b-a^2, a-b, b^2-ab) ,(b c-c a, c-a, a b-a^2):}| equals (a)a b c\ (b-c)(c-a)(a-b) (b) (b-c)(c-a)(a-b) (c) (a+b+c)(b-c)(c-a)(a-b) (d) none of these
Using properties of determinants, show the following: |((b+c)^2,ab,ca),(ab ,(a+c)^2,bc),(ac,bc,(a+b)^2)|=2abc(a+b+c)^3 - Sarthaks eConnect | Largest Online Education Community
![The determinant |{:(b^2-ab, b-c, bc-ac), (a b-a^2, a-b, b^2-ab) ,(b c-c a, c-a, a b-a^2):}| equals (a)a b c\ (b-c)(c-a)(a-b) (b) (b-c)(c-a)(a-b) (c) (a+b+c)(b-c)(c-a)(a-b) (d) none of these The determinant |{:(b^2-ab, b-c, bc-ac), (a b-a^2, a-b, b^2-ab) ,(b c-c a, c-a, a b-a^2):}| equals (a)a b c\ (b-c)(c-a)(a-b) (b) (b-c)(c-a)(a-b) (c) (a+b+c)(b-c)(c-a)(a-b) (d) none of these](https://d10lpgp6xz60nq.cloudfront.net/ss/web/731558.jpg)
The determinant |{:(b^2-ab, b-c, bc-ac), (a b-a^2, a-b, b^2-ab) ,(b c-c a, c-a, a b-a^2):}| equals (a)a b c\ (b-c)(c-a)(a-b) (b) (b-c)(c-a)(a-b) (c) (a+b+c)(b-c)(c-a)(a-b) (d) none of these
![In any triangle ABC, prove that `AB^2 + AC^2 = 2(AD^2 + BD^2)` , where D is the midpoint of BC. - YouTube In any triangle ABC, prove that `AB^2 + AC^2 = 2(AD^2 + BD^2)` , where D is the midpoint of BC. - YouTube](https://i.ytimg.com/vi/wu1SuC41ez4/maxresdefault.jpg)
In any triangle ABC, prove that `AB^2 + AC^2 = 2(AD^2 + BD^2)` , where D is the midpoint of BC. - YouTube
![SOLVED: P and Q are the points on the side BC of triangle ABC and AP=AQ prove that :AC+AB+BC is greater than 2AP +PQ SOLVED: P and Q are the points on the side BC of triangle ABC and AP=AQ prove that :AC+AB+BC is greater than 2AP +PQ](https://cdn.numerade.com/ask_previews/85447aea-613e-4130-85b6-88231d8d83ff_large.jpg)
SOLVED: P and Q are the points on the side BC of triangle ABC and AP=AQ prove that :AC+AB+BC is greater than 2AP +PQ
![Using properties of determinants, prove that: - a^2 ab ac | ab - b^2 bc | ca bc - c^2 = - 4a^2b^2c^2 Using properties of determinants, prove that: - a^2 ab ac | ab - b^2 bc | ca bc - c^2 = - 4a^2b^2c^2](https://haygot.s3.amazonaws.com/questions/2022326_1100710_ans_e24bfae4c6b6404aacce68b16bf09080.jpg)
Using properties of determinants, prove that: - a^2 ab ac | ab - b^2 bc | ca bc - c^2 = - 4a^2b^2c^2
![inequality - Prove that $(\frac{bc+ac+ab}{a+b+c})^{a+b+c} \geq \sqrt{(bc)^a( ac)^b(ab)^c}$ - Mathematics Stack Exchange inequality - Prove that $(\frac{bc+ac+ab}{a+b+c})^{a+b+c} \geq \sqrt{(bc)^a( ac)^b(ab)^c}$ - Mathematics Stack Exchange](https://i.stack.imgur.com/7VVVL.jpg)