Recall the formula `(abc)^2 = a^2 b^2 c^2 2(ab bc ca)` Given that `a^2 b^2 c^2 = 250 , ab bc ca = 3 ` Then we have `(abc)^2 = a^2 b^2 c^2 2 (ab bcca)`If a b c = 1, ab bc ca = 2 and abc = 3, then the value of a4 b4 c4 is equal to Byju's Answer Standard VIII Mathematics Factorisation by Regrouping Terms If a b c Question If Misc 13 Using properties of determinants, prove that 3a ab ac ba 3b bc ca cb 3c = 3 ( a b c) (ab bc ac) Taking LH S 3a ab ac ba 3b bc ca cb 3c Applying
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(ab+bc+ca)^3 formula-The following steps are followed while using (a 2 b 2 c 2 ) formula Firstly observe the pattern of the numbers whether the three numbers have ^2 as individual power or not Write down theHint ab bc ac = 2(abc)2−(a2b2c2) The number of ordered triples (a,b,c) of positive integers which satisfy the simultaneous equations ab bc = 44, ac bc = 33 Your solution is
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A3b3c3= (abc) a2b2c2−ab−bc−ac3abc Steve Duggan B Sc in Mathematics & Physics, University of Sydney Author has 15K answers and 48M answer views Updated 4 y A cube plus B cube plus C cube 3abc Ka Formula equal to a 3 b 3 c 3 = (a b c) (a 2 b 2 c 2 – ab – bc – ca) 3abc Which is an algebraic identity used to find the sum of ( a b b c c a) 3 = a b c ( a b c) 3 Prove that a, b, c must be terms of a G P I simplified this equation too ( a b) 3 ( b c) 3 ( c a) 3 = a b c ( a 3 b 3 c 3) I tried to subtract 3 ( a b c) 2 from both sides and it gave a factorised form
For nonzero numbers a a, b, b, and c c, \frac {\left (\frac {a} {b}\right)} {c} = \frac {a} {\left (\frac {b} {c}\right)} c(ba) = (cb)a Why some people say it's true Just like with multiplication, the order= (a b c)2– 2 (ab bc ca)=a2 b2 c2 So the formula is a2 b2 c2 = (a – b – c)2 2ab 2ac – 2bc Important points to be noted It's an algebraic formula that can factorize numbers AnYou can check the formulas of A plus B plus C Whole cube in three ways We are going to share the (abc)^3 algebra formulas for you as well as how to create (abc)^3 and proof we can write
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