[ベスト] (x y z)^3 identity 385116-(x+y+z)^3 identity

This video shows how to evaluate using the identity'x3y3z33xyz=(xyz)(x2y2z2xyyzzx)'To view more Educational content, please visit https//wwwyout

(x+y+z)^3 identity-(xyz)^3 (x y z)(x y z)(x y z) We multiply using the FOIL Method x * x = x^2 x * y = xy x * z = xz y * x = xy y * y = y^2 y * z = yz z * x = xz z * y = yz z * z = z^2 We now have x^2 xy5/6/17 (xy)^3 (yz)3 (zx)^3 = 3(xy)(yz)(zx) x3y3xy(12x2yy) y3z3yz(12y2zz) z3x3zx(12z2xx) = 3(xy)(yz)(zx) all the cubes then cancelxy(12x2yy) yz(12y

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In the expression, if we replace y with (− y), we will get the identity x 3 − y 318 x 3 – y 3 = (x – y)(x 2 xy y 2) 19 x 2 y 2 z 2 − xy – yz –zx = 1/2(x − y) 2 (y − z) 2 (z − x) 2 Algebraic Identities for Class 9 Maths

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