What is the value of frac{(1.2)^{3}+(0.8)^{3} | vedclass

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(B) The expression is in the form $\frac{a^3 + b^3 + c^3 - 3abc}{a^2 + b^2 + c^2 - ab - bc - ca}$,where $a = 1.2$,$b = 0.8$,and $c = 0.7$.We know the

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$2.7$

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(B) The expression is in the form $\frac{a^3 + b^3 + c^3 - 3abc}{a^2 + b^2 + c^2 - ab - bc - ca}$,where $a = 1.2$,$b = 0.8$,and $c = 0.7$.We know the algebraic identity: $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$.Substituting this into the numerator,the expression becomes: $\frac{(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)}{(a^2 + b^2 + c^2 - ab - bc - ca)}$.Canceling the common term,we get $a + b + c$.Substituting the values: $1.2 + 0.8 + 0.7 = 2.7$.

What is the value of $\frac{(1.2)^{3}+(0.8)^{3}+(0.7)^{3}-3(1.2)(0.8)(0.7)}{(1.2)^{2}+(0.8)^{2}+(0.7)^{2}-(1.2)(0.8)-(0.8)(0.7)-(0.7)(1.2)}$?

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