Answer
414.6k+ views
Hint: To find a cube of a number, we can simply multiply it with itself three times. But for finding a cube of a decimal, it is very tedious. So we expand the decimal into the sum or difference of an integer and the value after the decimal point. Then, we can use the formula for \[{(a + b)^3}\] and find the answer by comparatively easier multiplication and addition.
Complete step-by-step answer:
The given number is \[1.3\] , which is a decimal, so we can expand it as \[(1 + 0.3)\].
Now, instead of finding the cube of \[1.3\] , we can find the cube of \[(1 + 0.3)\] which will have the same value.
The cube of \[(1 + 0.3)\] can be found using the formula of \[{(a + b)^3}\].
Thus,
\[
\Rightarrow {(a + b)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3} \\
\Rightarrow {(1 + 0.3)^3} = {(1)^3} + 3{(1)^2}(0.3) + 3(1){(0.3)^2} + {(0.3)^3} \\
\Rightarrow {(1 + 0.3)^3} = 1 + 3(1)(0.3) + 3(1)(0.09) + (0.027) \\
\Rightarrow {(1 + 0.3)^3} = 1 + 0.9 + 0.27 + 0.027 \\
\Rightarrow {(1 + 0.3)^3} = 2.197 \\
\]
So, the cube of \[1.3\] comes out to be \[2.197\].
Hence, option (D) is correct.
Note: Avoid writing \[1.3\] as \[(2 - 0.7)\] , as that will unnecessarily complicate the calculations. Also the student may get confused between addition and subtraction signs.
Another simpler way of solving this question could be by eliminating the options. As \[1.3\] is greater than \[1\] and less than \[2\] , so its cube will also be greater than \[{1^3}\] and less than \[{2^3}\] . Which means the cube of \[1.3\] will be more than \[1\] and less than \[8\] , which is possible only in option (D). But once again this can be done only if options are given with large differences, so students should refrain from this.
Another, much more precise method can be by converting the decimal into a fraction, then find the cube of both numerator and denominator, which will give the same result .
The decimal can be written into fraction as \[\dfrac{{1.3}}{1} = \dfrac{{13}}{{10}}\] . Now we can cube both the numerator and denominator as \[{(\dfrac{{13}}{{10}})^3} = \dfrac{{{{13}^3}}}{{{{10}^3}}} = \dfrac{{2197}}{{1000}} = 2.197\].
But, here too there are a lot of calculations involved which makes the chances of making mistakes much higher.
Complete step-by-step answer:
The given number is \[1.3\] , which is a decimal, so we can expand it as \[(1 + 0.3)\].
Now, instead of finding the cube of \[1.3\] , we can find the cube of \[(1 + 0.3)\] which will have the same value.
The cube of \[(1 + 0.3)\] can be found using the formula of \[{(a + b)^3}\].
Thus,
\[
\Rightarrow {(a + b)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3} \\
\Rightarrow {(1 + 0.3)^3} = {(1)^3} + 3{(1)^2}(0.3) + 3(1){(0.3)^2} + {(0.3)^3} \\
\Rightarrow {(1 + 0.3)^3} = 1 + 3(1)(0.3) + 3(1)(0.09) + (0.027) \\
\Rightarrow {(1 + 0.3)^3} = 1 + 0.9 + 0.27 + 0.027 \\
\Rightarrow {(1 + 0.3)^3} = 2.197 \\
\]
So, the cube of \[1.3\] comes out to be \[2.197\].
Hence, option (D) is correct.
Note: Avoid writing \[1.3\] as \[(2 - 0.7)\] , as that will unnecessarily complicate the calculations. Also the student may get confused between addition and subtraction signs.
Another simpler way of solving this question could be by eliminating the options. As \[1.3\] is greater than \[1\] and less than \[2\] , so its cube will also be greater than \[{1^3}\] and less than \[{2^3}\] . Which means the cube of \[1.3\] will be more than \[1\] and less than \[8\] , which is possible only in option (D). But once again this can be done only if options are given with large differences, so students should refrain from this.
Another, much more precise method can be by converting the decimal into a fraction, then find the cube of both numerator and denominator, which will give the same result .
The decimal can be written into fraction as \[\dfrac{{1.3}}{1} = \dfrac{{13}}{{10}}\] . Now we can cube both the numerator and denominator as \[{(\dfrac{{13}}{{10}})^3} = \dfrac{{{{13}^3}}}{{{{10}^3}}} = \dfrac{{2197}}{{1000}} = 2.197\].
But, here too there are a lot of calculations involved which makes the chances of making mistakes much higher.
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