If the zeroes of the polynomial x3 – 3x2 + x + 1 are a-b, a, a + b

1.	If the zeroes of the polynomial x3 – 3x2 + x + 1 are a-b, a, a + b, find a and b.❌ no spam pleach ♡
Answer» ᴄᴏʀʀᴇᴄᴛ qᴜᴇꜱᴛɪᴏɴ: The ZEROES of the polynomial x³ – 3x² + x + 1 are— a-B, a, a + b ᴛᴏ ꜰɪɴᴅ: A and B ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ ★For a Cubic Polynomial: $~~~~~~~~~$ p(x) = ax³ + bx² + CX + d ★With zeroes $\alpha$ , $\beta$ and $\gamma$ We have, $\alpha$ + $\beta$ + $\gamma$ = $\frac{ - b}{a}$ $\alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a}$ $\alpha \beta \gamma = \frac{ - d}{a}$ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ NOW- $~~~~~~~~~$ p(x) = x³ - 3x² + x + 1 ★Comparing with p(x) = Ax³ + Bx² + Cx + D A = 1 B = -3 C = 1 D = 1 Zeroes are— $\alpha = a - b$ $\beta = a$ $\gamma = ab$ $\\ \\$ ★Sum Of Zeroes— $~~~~~~~~~$ $\implies$ Sum of zeroes = $\frac{ - b}{a}$ $~~~~~~~~~$ $\implies$ $\alpha + \beta + \gamma = \frac{ - b}{a}$ $~~~~~~~~~$ $\implies$ a - b + a + a + b = 3 $~~~~~~~~~$ $\implies$ 3a = 3 $~~~~~~~~~$ $\implies$ a = 1 $\\ \\$ ★Sum Of Product Zeroes— Sum of product zeroes = $\frac{c}{a}$ $\implies$ $\alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a}$ $\\ \\$ ⟶ (a-b)a + a(a+b) + (a+b) (a-b) = 1 ⟶ a² - ba + a² + ab + a² - b² = 1 ⟶ a² + a² + a² - b² = 1 ⟶ 3a² - b² = 1 PUTTING A=1 ⟶ 3(1)² - b² = 1 ⟶ 3 - b² = 1 ⟶ 3 - a = b² ⟶ b² = 2 ⟶ b = $± \sqrt{2}$ $\\ \\$ Therefore, a = 1 & b = $± \sqrt{2}$ _____________________________________