Question

Evaluate $ {{\left( {{i}^{18}}+{{\left( \dfrac{1}{i} \right)}^{25}} \right)}^{3}} $ .

Answer 1

Hint: In this question, we have to evaluate an expression given in terms of i. For this, we will use the value of i which is $ \sqrt{-1} $ . We will first find the value of $ {{i}^{18}} $ and then find the value of $ {{\left( \dfrac{1}{i} \right)}^{25}} $ . After that, we will use the property of $ {{\left( a+b \right)}^{3}} $ to finally evaluate our answer $ {{\left( a+b \right)}^{3}} $ is equal to $ {{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}} $ . We will also use property of exponent $ {{a}^{m}}\cdot {{a}^{n}}={{a}^{m+n}} $ to evaluate our answer.

Complete step by step answer:
Here we are given the expression as $ {{\left( {{i}^{18}}+{{\left( \dfrac{1}{i} \right)}^{25}} \right)}^{3}} $ .
For this, let us first find the values of $ {{i}^{18}}\text{ and }{{\left( \dfrac{1}{i} \right)}^{25}} $ separately.
For $ {{i}^{18}} $ .
We know that the value is given as $ i=\sqrt{-1} $ .
Squaring both sides we get $ {{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}\Rightarrow {{i}^{2}}=-1 $ .
Hence the value of $ {{i}^{2}}=-1 $ .
Again squaring both sides we get: $ {{\left( {{i}^{2}} \right)}^{2}}={{\left( -1 \right)}^{2}}\Rightarrow {{i}^{4}}=1 $ .
Taking the power as 4 on both sides we get: $ {{\left( {{i}^{4}} \right)}^{4}}={{\left( 1 \right)}^{4}}\Rightarrow {{i}^{16}}=1 $ .
Now we need to evaluate $ {{i}^{18}} $ so let us multiply both sides by $ {{i}^{2}} $ we get: $ {{i}^{16}}\cdot {{i}^{2}}=1\cdot {{i}^{2}} $ .
We know that, $ {{x}^{m}}\cdot {{x}^{n}}={{x}^{m+n}} $ so we get: $ {{i}^{16+2}}={{i}^{2}}\Rightarrow {{i}^{18}}={{i}^{2}} $ .
Now we know that $ {{i}^{2}}=-1 $ so value of $ {{i}^{18}} $ becomes equal to -1.
Therefore, $ {{i}^{18}}=-1 $ .
For $ {{\left( \dfrac{1}{i} \right)}^{25}} $ .
Let us first simplify $ \dfrac{1}{i} $ so that we do not have i in the denominator. Multiplying the numerator and the denominator by i we get $ \dfrac{1}{{{i}^{2}}} $ . Now, the value of $ {{i}^{2}} $ is -1 so we get: $ \dfrac{1}{i}=-i $ .
So $ {{\left( \dfrac{1}{i} \right)}^{25}} $ becomes equal to $ {{\left( -i \right)}^{25}} $ .
Odd powers do not cancel negative signs, so it can be written as $ -{{i}^{25}} $ .
Now we know that $ {{i}^{4}}=1 $ so taking the power 6 on both sides we get: $ {{\left( {{i}^{4}} \right)}^{6}}={{\left( 1 \right)}^{6}}\Rightarrow {{i}^{24}}=1 $ .
Multiplying both sides by i, we get: $ {{i}^{24}}\cdot i=1\cdot i $ .
Using $ {{a}^{m}}\cdot {{a}^{n}}={{a}^{m+n}} $ we get: \[{{i}^{24+1}}=1\cdot i\Rightarrow {{i}^{25}}=i\].
Hence the value of $ {{\left( \dfrac{1}{i} \right)}^{25}} $ become equal to \[-{{i}^{25}}=-i\].
Now putting the values of $ {{i}^{18}}\text{ and }{{\left( \dfrac{1}{i} \right)}^{25}} $ in the given expression we get:
 $ {{\left( {{i}^{18}}+{{\left( \dfrac{1}{i} \right)}^{25}} \right)}^{3}}={{\left( -1-i \right)}^{3}} $ .
Taking negative sign common we get:
 $ {{\left( {{i}^{18}}+{{\left( \dfrac{1}{i} \right)}^{25}} \right)}^{3}}={{\left( -\left( 1+i \right) \right)}^{3}} $ .
Odd powers do not cancel negative signs so it can be written as $ -{{\left( 1+i \right)}^{3}} $ .
We know that $ {{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}} $ so we get: $ -\left[ {{\left( 1 \right)}^{3}}+{{\left( i \right)}^{3}}+3{{\left( 1 \right)}^{2}}\left( i \right)+3\left( 1 \right){{\left( i \right)}^{2}} \right] $ .
We know that $ {{i}^{2}}=-1 $ multiplying i on both sides we get: $ {{i}^{3}}=-i $ .
Putting value of $ {{i}^{2}}\text{ and }{{i}^{3}} $ in the given expression we get: $ -\left[ 1-i+3i-3 \right]\Rightarrow -\left[ -2+2i \right]=2-2i $ .
Hence our required value of the expression $ {{\left( {{i}^{18}}+{{\left( \dfrac{1}{i} \right)}^{25}} \right)}^{3}} $ is equal to $ 2-2i $ .

Note:
Students should take care of the signs while solving these sums. They can also remember the values of $ {{i}^{2}}\text{ and }{{i}^{4}} $ . The student should note that $ {{i}^{4m}} $ is always equal to 1 form being any integer.