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The value of (0.98)³ + (0.02)³ + 3 * 0.98 * 0.02 - 1 = ?

11.98
21.09
31
40

Answer:- 1

Explanation:-

Solution:

\begin{aligned} According to question \\ {(0.98)}^{3} + {(0.02)}^{3} + 3 \times 0.98 \times 0.02 - 1 \\ = {(0.98)}^{3} + {(0.02)}^{3} + 3 \times 0.98 \times 0.02 (0.98 + 0.02) - 1 \\ = {(0.98 + 0.02)}^{3} - 1 [∵ {(a + b)}^{3} = a^{3} + b^{3} + 3 a b (a + b)] \\ = {(1)}^{3} - 1 \\ = 0 \\ Alternate \\ a = 0.98 \\ b = 0.02 \\ c = - 1 \\ a + b + (- c) = 0 \\ S o, \\ a^{3} + b^{3} + (- c)^{3} - 3 a b c \\ = 0 \end{aligned}

$\eqalign{ & {\text{According to question}} \cr & \,{\left( {0.98} \right)^3} + {\left( {0.02} \right)^3} + 3 \times 0.98 \times 0.02 - 1 \cr & = {\left( {0.98} \right)^3} + {\left( {0.02} \right)^3} + 3 \times 0.98 \times 0.02\left( {0.98 + 0.02} \right) - 1 \cr & = {\left( {0.98 + 0.02} \right)^3} - 1\,\,\,\left[ {\because {{\left( {a + b} \right)}^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)} \right] \cr & = {\left( 1 \right)^3} - 1 \cr & = 0 \cr & {\text{Alternate}} \cr & a = 0.98 \cr & b = 0.02 \cr & c = - 1 \cr & a + b + \left( { - c} \right) = 0 \cr & So, \cr & {a^3} + {b^3} + {( - c)^3} - 3abc \cr & = 0 \cr}$

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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} According to question \\ {(0.98)}^{3} + {(0.02)}^{3} + 3 \times 0.98 \times 0.02 - 1 \\ = {(0.98)}^{3} + {(0.02)}^{3} + 3 \times 0.98 \times 0.02 (0.98 + 0.02) - 1 \\ = {(0.98 + 0.02)}^{3} - 1 [∵ {(a + b)}^{3} = a^{3} + b^{3} + 3 a b (a + b)] \\ = {(1)}^{3} - 1 \\ = 0 \\ Alternate \\ a = 0.98 \\ b = 0.02 \\ c = - 1 \\ a + b + (- c) = 0 \\ S o, \\ a^{3} + b^{3} + (- c)^{3} - 3 a b c \\ = 0 \end{aligned}

$\eqalign{ & {\text{According to question}} \cr & \,{\left( {0.98} \right)^3} + {\left( {0.02} \right)^3} + 3 \times 0.98 \times 0.02 - 1 \cr & = {\left( {0.98} \right)^3} + {\left( {0.02} \right)^3} + 3 \times 0.98 \times 0.02\left( {0.98 + 0.02} \right) - 1 \cr & = {\left( {0.98 + 0.02} \right)^3} - 1\,\,\,\left[ {\because {{\left( {a + b} \right)}^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)} \right] \cr & = {\left( 1 \right)^3} - 1 \cr & = 0 \cr & {\text{Alternate}} \cr & a = 0.98 \cr & b = 0.02 \cr & c = - 1 \cr & a + b + \left( { - c} \right) = 0 \cr & So, \cr & {a^3} + {b^3} + {( - c)^3} - 3abc \cr & = 0 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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