X39x2y27xy227y3 Final result (x 3y)3 Step by step solution Step 1 Equation at the end of step 1 (((x3)((9•(x2))•y))(27x•(y2)))33y3 Step 2 Equation at the end of step 2 How toAnswer (1 of 9) x^2 y^2 => x^2 (1)y^2 => x^2 i^2 *y^2 ( We know i^2= 1) => (x)^2 (iy)^2 => (xiy)(xiy) Answerx^2y^2z^2)^2 4x^2y^2 ( (x^2 y^2 z^2 2xy ) ( x^2 y^2 z^2 2xy ) ( x^2 2xy y^2 z^2 ) ( x^2 2xy y^2 z^2 stronkboi152 stronkboi152
How To Factorise A Polynomial By Splitting The Middle Term A Plus Topper
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1 Without using mathematical tables or a calculator, evaluate #(28 times )/(0645 times 2175)# 2 a)Expand the expression #(3x7)(2x7)# b)Without using long multiplication or a calculator, use the above expansion with an appropriate Let x, y, z > 0 satisfy x y y z z x 2 x y z = 1 Prove that 4 x y z ≥ 2 The condition invites the factoring ( 1 x) ( 1 y) ( 1 z) x y z − 2 = x y z, but having the factor 4 in the desired inequality makes things more difficult algebraprecalculus inequality Share asked at 1013 nan 1,035 5 16 Add a comment
