いろいろ (a+b+c)3 ka formula 163009-(a+b+c)2 ka formula

Using determinants, we can write the result as a × b = a 2 a 3 b 2 b 3 i − a 1 a 3 b 1 b 3 j a 1 a 2 b 1 b 2 k Looking at the formula for the 3 × 3 determinant, we see that the formula for a cross product looks a lot like the formula for the 3 × 3 determinantExample 3 A = {a, b, c, d, e}, B = {x, y, z} and C = {a, e, x} find the value of n(AUBUC) Solution n(AUBUC)=n(A)n(B) n(C)n(AnB)n(BnC)n(CnA)n(AnBnC) n (A) = 5 n (B) = 3 n (C) = 3 A n B = {a, b, c, d, e} n {x, y, z} = { } n (A n B) = 0 B n C = {x, y, z} n {a, e, x} = {x} n (B n C) = 1The Triangle Formula are given below as, Perimeter of a triangle = a b c \Area\;

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(a+b+c)2 ka formula

(a+b+c)2 ka formula-A 3 b 3 = (a b) (a 2 b 2 − ab) (a b c) 3 = a 3 b 3 c 3 3(a b)(b c)(c a) a 3 b 3 c 3 − 3abc = (a b c) (a 2 b 2 c 2 − ab − bc − ac)Solving for B means to isolate it to one side of the equation, doesn't matter which side A= ABC/3 Start by subtracting A from both sides

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MATHEMATICAL FORMULAE Algebra MATHEMATICAL FORMULAE Algebra 1 (ab)2=a22abb2;a2b2=(ab)2−2ab 2 (a−b)2=a2−2abb;a2b2=(a−b)22ab 3 (abc)2=a2b2c22(abbcca) 4 (ab)3=a3b33ab(ab);a3b3=(ab)−3ab(ab) 5(a b c) 2 = a b (c) 2 = a 2 b 2 (c) 2 2ab 2 (b) (c) 2 (c) (a) = a 2 b 2 c 2 2ab – 2bc 2ca Therefore, (a b c) 2 = a 2 b 2 c 2 2ab – 2bc 2ca (a b c) 2 = a ( b) c 2 = a 2 (b 2) c 2 2 (a) (b) 2 (b) (c) 2 (c) (a) = a 2 b 2 c 2 – 2ab – 2bc 2ca Therefore, (a b c) 2 = a 2 b 2 c 2 – 2ab – 2bc 2ca(a b c)³ = a³ b³ c³ 3 (a b) (b c) (a c) It can be written as (a b c)³ a³ b³ c³ = 3 (a b) (b c) (a c) (1) Consider the LHS of equation (1), (a b c)³ a³ b³ c³ = a³ b³ c³ 3 ab (a b) 3 bc (b c) 3 ac (a c) 6 abc a³ b³ c³

The formula for Solution To find the formula for Multiply twice and simplify the terms Squaring the whole equation, we can see that the value of (abc)^{2} is Taking 2 common in we get Therefore, the formula for is This formula is used to express the algebraic identity if they are 3 terms given according to the operations involvedThe Triangle Formula are given below as, Perimeter of a triangle = a b c \Area\;(a b) 3 = a 3 b 3 3ab(a b) (a – b) 3 = a 3 – 3a 2 b 3ab 2 – b 3 = a 3 – b 3 – 3ab(a – b) a 3 – b 3 = (a – b)(a 2 ab b 2) a 3 b 3 = (a b)(a 2 – ab b 2) (a b) 4 = a 4 4a 3 b 6a 2 b 2 4ab 3 b 4

A 3 b 3 c 3 3abc = (a b c)(a 2 b 2 c 2 ab bc ca) Where a = 2a, b = 3b and c = 5c Now apply values of a, b and c on the LHS of identity ie a 3 b 3 c 3 3abc = (a b c)(a 2 b 2 c 2 ab bc ca) and we get = (2a 3b 5c) { (2a) 2 (3b) 2 (5c) 2 – (2a)(3b) – (3b)(5c) – (5c)(2a) } Expand the exponential forms and we getLooking at the formula for the $3 \times 3$ determinant, we see that the formula for a cross product looks a lot like the formula for the $3 \times 3$ determinant If we allow a matrix to have the vector $\vc{i}$, $\vc{j}$, and $\vc{k}$ as entries (OK, maybe this doesn't make sense, but this is just as a tool to remember the cross product), theOnly the use of the quadratic formula, as well as the basics of completing the square will be discussed here (since the derivation of the formula involves completing the square) Below is the quadratic formula, as well as its derivation Derivation of the Quadratic Formula From this point, it is possible to complete the square using the relationship that x 2 bx c = (x h) 2 k Continuing the derivation using this relationship

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There are mainy two forms to write the formula of a 3 b 3 c 3 First one is a 3 b 3 c 3 = ( a b c) a 2 b 2 c 2 − a b − b c − a c 3 a b c Second one is a 3 b 3 c 3 = ( a b c) 1 2 ( a − b) 2 ( b − c) 2 ( c − a) 2 3 a b c(a b c) 2 = a b (c) 2 = a 2 b 2 (c) 2 2ab 2 (b) (c) 2 (c) (a) = a 2 b 2 c 2 2ab – 2bc 2ca Therefore, (a b c) 2 = a 2 b 2 c 2 2ab – 2bc 2ca (a b c) 2 = a ( b) c 2 = a 2 (b 2) c 2 2 (a) (b) 2 (b) (c) 2 (c) (a) = a 2 b 2 c 2 – 2ab – 2bc 2ca Therefore, (a b c) 2 = a 2 b 2 c 2 – 2ab – 2bc 2ca(abc) 3 a 3 b 3 c 3 We can choose three "a"'s for the cube in one way C(3,3)=1, or we can choose an a from the first factor and one from the second and one from the third, being the only way to make a3 The coefficient of the cubes is therefore 1 (It's the same for a, b and c, of course) 3a 2 b3a 2 c Next, we consider the a 2 terms We can choose two a's from 3 factors in C(3,2) ways=3

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A= ABC/3 Answer by nerdybill (7384) ( Show Source ) You can put this solution on YOUR website!As stated in the title, I'm supposed to show that (a b c) 3 = a 3 b 3 c 3 (a b c) (a b a c b c)Only the use of the quadratic formula, as well as the basics of completing the square will be discussed here (since the derivation of the formula involves completing the square) Below is the quadratic formula, as well as its derivation Derivation of the Quadratic Formula From this point, it is possible to complete the square using the relationship that x 2 bx c = (x h) 2 k Continuing the derivation using this relationship

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Video Lesson Khan Academy Video Quadratic Formula 1;Binomial Theorem (ab)1 = a b ( a b) 1 = a b (ab)2 = a2 2abb2 ( a b) 2 = a 2 2 a b b 2 (ab)3 = a3 3a2b 3ab2 b3 ( a b) 3 = a 3 3 a 2 b 3 a b 2 b 3 (ab)4 = a4 4a3b 6a2b2 4ab3 b4 ( a b) 4 = a 4 4 a 3 b 6 a 2 b 2 4 a b 3 b 4Example 3 A = {a, b, c, d, e}, B = {x, y, z} and C = {a, e, x} find the value of n(AUBUC) Solution n(AUBUC)=n(A)n(B) n(C)n(AnB)n(BnC)n(CnA)n(AnBnC) n (A) = 5 n (B) = 3 n (C) = 3 A n B = {a, b, c, d, e} n {x, y, z} = { } n (A n B) = 0 B n C = {x, y, z} n {a, e, x} = {x} n (B n C) = 1

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(a b c) 2 = a 2 b 2 c 2 2ab 2bc 2ca (a – b – c) 2 = a 2 b 2 c 2 – 2ab 2bc – 2ca (a b) 3 = a 3 3a 2 b 3ab 2 b 3;Triangle= \frac{1}{2}bh\ Where, b is the base of the triangle h is the height of the triangle If only 2 sides and an internal angle is given then the remaining sides and angles can be calculated using the below formulaSolve an equation of the form a x 2 b x c = 0 by using the quadratic formula x = − b ± √ b 2 − 4 a c 2 a StepByStep Guide Learn all about the quadratic formula with this stepbystep guide Quadratic Formula, The MathPapa Guide;

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