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If a = 4.36, b = 2.39 and c = 1.97, then the value of a³ - b³ - c³ - 3abc is?

13.94
22.39
30
41

Answer:- 1

Explanation:-

Solution:

\begin{aligned} a = 4.36 \\ b = 2.39 \\ c = 1.97 \\ a - b - c \\ = 4.36 - 2.39 - 1.97 \\ = 0 \\ a^{3} - b^{3} - c^{3} - 3 a b c \\ = \frac{1}{2} (a - b - c) [{(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2}] \\ = 0 \end{aligned}

$\eqalign{ & a = 4.36 \cr & b = 2.39 \cr & c = 1.97 \cr & {\text{ }}a - b - c \cr & = 4.36 - 2.39 - 1.97 \cr & = 0 \cr & {\text{ }}{{\text{a}}^3} - {b^3} - {c^3} - 3abc \cr & = \frac{1}{2}\left( {a - b - c} \right)\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] \cr & = 0 \cr}$

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", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} a = 4.36 \\ b = 2.39 \\ c = 1.97 \\ a - b - c \\ = 4.36 - 2.39 - 1.97 \\ = 0 \\ a^{3} - b^{3} - c^{3} - 3 a b c \\ = \frac{1}{2} (a - b - c) [{(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2}] \\ = 0 \end{aligned}

$\eqalign{ & a = 4.36 \cr & b = 2.39 \cr & c = 1.97 \cr & {\text{ }}a - b - c \cr & = 4.36 - 2.39 - 1.97 \cr & = 0 \cr & {\text{ }}{{\text{a}}^3} - {b^3} - {c^3} - 3abc \cr & = \frac{1}{2}\left( {a - b - c} \right)\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] \cr & = 0 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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