Question

If the expression ${\left( {1 + ir} \right)^3}$ is of the form $s\left( {1 + i} \right)$ for possible real s and r is also real and, then the sum of all possible values for r is?

Answer 1

Hint: First assume a variable ‘a’ and take it as ir. Now solve the cubic equation. Equate it to the given term. From this you get two relations between s,r. Try to eliminate s and get a relation of r. This will be a cubic equation. By using general algebra we know the sum of roots of equation $a{x^3} + b{x^2} + cx + d$ is given by $\dfrac{{ - b}}{a}$. Using this find the sum of all possible values of r, which is the required result.

Complete step-by-step solution-
Given cubic term in the question is written as:
${\left( {1 + ir} \right)^3}$
Let us assume that ir as a variable denoted by a:
$a = ir$ ………….(1)
By substituting this into the cubic term, we get:
${\left( {1 + a} \right)^3}$
We can write the above term, in the form of:
\[\left( {1 + a} \right)\left( {1 + a} \right)\left( {1 + a} \right)\]
Distributive law says that \[\left( {a + b} \right).c = ac + bc\]
By applying distributive law on the first two terms, we get:
\[\left( {1 + a + a + {a^2}} \right)\left( {1 + a} \right)\]
By simplifying the first term, we get it in form of:
\[\left( {1 + 2a + {a^2}} \right)\left( {1 + a} \right)\]
The above terms can be written in the form of:
\[\left( {{a^2} + 2a + 1} \right)\left( {a + 1} \right)\]
By applying the distributive law again, we can say that:
\[{a^3} + 2{a^2} + a + {a^2} + 2a + 1\]
By bringing some degree terms together we get it as:
 \[{a^3} + \left( {2{a^2} + {a^2}} \right) + \left( {2a + a} \right) + 1\]
By, simplifying the term, we get it as:
\[{a^3} + 3{a^2} + 3a + 1\]
By substituting the equation (1) back into this term, we get:
\[{\left( {ir} \right)^3} + 3{\left( {ir} \right)^2} + 3\left( {ir} \right) + 1\]
By simplifying all the terms, we get the term as:
\[{r^3}{i^3} + 3{r^2}{i^2} + 3ri + 1\]
By substituting\[{i^3} = - i\], \[{i^2} = - 1\] , we get the term as:
\[ - {r^3}i - 3{r^2} + 3ri + 1\]
Given in the question this term is equal to \[{\rm{s}} + {\rm{si}}\].
\[ - 3{r^2} + 1 - {r^3}i + 3ri = s + si\]
By equating the term correspondingly we get it as:
\[1 - 3{r^2} = s\,\,;\,\,3r - {r^3} = s\]
By eliminating s from both equations, we get: \[1 - 3{r^2} = \,\,3r - {r^3}\]
Be bringing all terms at one side we \[{r^3} - 3{r^2} - 3r + 1 = \,\,0\]
We know in $a{x^3} + b{x^2} + cx + d$, sum of roots is $\dfrac{{ - b}}{c}$
Here $a = 1$. $b = - 3$, so sum of roots is given by:
Sum of all possible $r = \dfrac{{ - \left( { - 3} \right)}}{1} = 3$
Therefore, 3 is the value of the required expression asked in question.

Hint: Alternate method is to use algebra identity ${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)$ to the polynomial directly and get till S’s relation in single step. As we have s directly in both equations we eliminate directly, if not we must use any suitable algebraic operation. Be careful with “-“ while comparing terms.