Q60.If x= 1+ (cube root 5) + (cube root 25), find the value of (x cube)-3(x square)-12x+6.

We are given:

x = 1 + ∛5 + ∛25

We need to find the value of:

x³ - 3x² - 12x + 6

Step 1: Identify the relation between the terms
Notice that ∛25 = (∛5)². So let:

a = ∛5

Then:

x = 1 + a + a²

Since the cube roots follow the identity:

a³ = 5

Step 2: Cube the given expression
By the identity:

x³ = (1 + a + a²)³
Expanding using the identity for cubic expansion:

x³ = 1³ + a³ + (a²)³ + 3(1)(a)(a²) + 3(1)(a²) + 3(a)(1)

Since a³ = 5, this becomes:

x³ = 1 + 5 + 25 + 3(a²) + 3a + 3a²
x³ = 31 + 6a + 6a²

Since a + a² = x - 1,

x³ = 31 + 6(x - 1)
x³ = 31 + 6x - 6
x³ = 6x + 25

Step 3: Solve for the required expression
We need to evaluate:

x³ - 3x² - 12x + 6

From earlier,

x³ = 6x + 25

Now,

x³ - 3x² - 12x + 6
= (6x + 25) - 3x² - 12x + 6
= -3x² - 6x + 31

Step 4: Final Answer
Final Answer: 31 - 3x² - 6x