Question

In all of this question let (*) represent the equation 3x³ + 9x² + 18x - 96 = 0.

(a) Carry out the substitution x = y - (b/3a) to transform (*) into the form y³ + my = n.

(b) Apply Cardano's formula to find one real solution to the equation in Part (a) and simplify your answer. State the corresponding solution to (*).

(c) Show that the other solutions to (*) are not real numbers.

[Please show your detailed work]

Answer 1

Given equation is : 3x³+9x²+18x-96 = 0

a) Here, b = 9 and a = 3.

Now, we substitute x = y-[9/(3*3)] ,i.e., x = y-1 in the given equation we get,

3(y-1)³+9(y-1)²+18(y-1)-96 = 0

i.e., 3[(y-1)³+3(y-1)²+3(y-1)+1]+9(y-1)-99 = 0

i.e., 3(y-1+1)³+9y-9-99 = 0

i.e., 3y³+9y-108 = 0

i.e., y³+3y-36 = 0

Therefore, the equation becomes, y³+3y-36 = 0..................(i).

b) Let y = u+v.

Then, y³ = u³+v³+3uv(u+v)

i.e., y³ = u³+v³+3uvy

i.e., y³-3uvy-(u³+v³) = 0.................(ii)

Comparing (i) and (ii) we get,

uv = -1 and u³+v³ = 36

Now, u³ = ,i.e., u³ =

i.e., u³ =

And, v³ =

If p denotes any one of the three values of , then the three values of u are p, p, ²p where is an imaginary cube root of unity.

And, since uv = -1, the three corresponding values of v are -1/p, -²/p, -/p.

Hence the values of y are , , and values of x are p-(1/p)-1, p-(²/p)-1, ²p-(/p)-1.

Here, only p-(1/p)-1 is the only real solution.

c) The other solutions contains which is an imaginary number.

Therefore, other solutions are not real numbers.