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Question: If the expression \({\left( {1 + ir} \right)^3}\) is of the form \(s\left( {1 + i} \right)\) for pos...

If the expression (1+ir)3{\left( {1 + ir} \right)^3} is of the form s(1+i)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?

Explanation

Solution

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 ax3+bx2+cx+da{x^3} + b{x^2} + cx + d is given by −ba\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:
(1+ir)3{\left( {1 + ir} \right)^3}
Let us assume that ir as a variable denoted by a:
a=ira = ir ………….(1)
By substituting this into the cubic term, we get:
(1+a)3{\left( {1 + a} \right)^3}
We can write the above term, in the form of:
(1+a)(1+a)(1+a)\left( {1 + a} \right)\left( {1 + a} \right)\left( {1 + a} \right)
Distributive law says that (a+b).c=ac+bc\left( {a + b} \right).c = ac + bc
By applying distributive law on the first two terms, we get:
(1+a+a+a2)(1+a)\left( {1 + a + a + {a^2}} \right)\left( {1 + a} \right)
By simplifying the first term, we get it in form of:
(1+2a+a2)(1+a)\left( {1 + 2a + {a^2}} \right)\left( {1 + a} \right)
The above terms can be written in the form of:
(a2+2a+1)(a+1)\left( {{a^2} + 2a + 1} \right)\left( {a + 1} \right)
By applying the distributive law again, we can say that:
a3+2a2+a+a2+2a+1{a^3} + 2{a^2} + a + {a^2} + 2a + 1
By bringing some degree terms together we get it as:
a3+(2a2+a2)+(2a+a)+1{a^3} + \left( {2{a^2} + {a^2}} \right) + \left( {2a + a} \right) + 1
By, simplifying the term, we get it as:
a3+3a2+3a+1{a^3} + 3{a^2} + 3a + 1
By substituting the equation (1) back into this term, we get:
(ir)3+3(ir)2+3(ir)+1{\left( {ir} \right)^3} + 3{\left( {ir} \right)^2} + 3\left( {ir} \right) + 1
By simplifying all the terms, we get the term as:
r3i3+3r2i2+3ri+1{r^3}{i^3} + 3{r^2}{i^2} + 3ri + 1
By substitutingi3=−i{i^3} = - i, i2=−1{i^2} = - 1 , we get the term as:
−r3i−3r2+3ri+1- {r^3}i - 3{r^2} + 3ri + 1
Given in the question this term is equal to s+si{\rm{s}} + {\rm{si}}.
−3r2+1−r3i+3ri=s+si- 3{r^2} + 1 - {r^3}i + 3ri = s + si
By equating the term correspondingly we get it as:
1−3r2=s  ;  3r−r3=s1 - 3{r^2} = s\,\,;\,\,3r - {r^3} = s
By eliminating s from both equations, we get: 1−3r2=  3r−r31 - 3{r^2} = \,\,3r - {r^3}
Be bringing all terms at one side we r3−3r2−3r+1=  0{r^3} - 3{r^2} - 3r + 1 = \,\,0
We know in ax3+bx2+cx+da{x^3} + b{x^2} + cx + d, sum of roots is −bc\dfrac{{ - b}}{c}
Here a=1a = 1. b=−3b = - 3, so sum of roots is given by:
Sum of all possible r=−(−3)1=3r = \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 (a+b)3=a3+b3+3ab(a+b){\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.