Given that the zeroes of \(x^3-6x^2+3x+10\) are of the form \(a,\, a+b,\, a+2b\), find \(a\), \(b\) and the zeroes.
\(a=5,\; b=-3\) (equivalently \(a=-1,\; b=3\)); Zeroes: \(\{5,\,2,\,-1\}\).
Step 1: Use Vieta’s formulas (sum and product of zeroes).
For \(x^3 - 6x^2 + 3x + 10\):
Sum of zeroes = \(6\).
Product of zeroes = \(-10\).
Step 2: Express these using \(a, b\).
Sum: \(a + (a+b) + (a+2b) = 3a + 3b = 6\).
So, \(a + b = 2\).
Product: \(a(a+b)(a+2b) = -10\).
Step 3: Substitute \(b = 2 - a\) into the product.
\(a\cdot 2 \cdot (4 - a) = -10\) (since \(a+2b = a + 2(2-a) = 4 - a\)).
So, \(2a(4 - a) = -10\Rightarrow a(4 - a) = -5\).
\(\Rightarrow -a^2 + 4a + 5 = 0 \Rightarrow a^2 - 4a - 5 = 0\).
Hence, \(a = 5\) or \(a = -1\).
Then \(b = 2 - a\Rightarrow b = -3\) or \(b = 3\).
Step 4: Write the zeroes.
For \(a=5, b=-3\): zeroes are \(5, 2, -1\).
For \(a=-1, b=3\): zeroes are \(-1, 2, 5\) (same set).