Solution :
Let y = x + \(x^2\). Then,
\((1 + x + x^2)^3\) = \((1 + y)^3\)
= \(^3C_0\) + \(^3C_1 y\) + \(^3C_2 y^2\) + \(^3C_3 y^3\)
= \(1 + 3y + 3y^2 + y^3\) = 1 + 3\((x + x^2)\) + 3\((x + x^2)^2\) + \((x + x^2)^3\)
= \(x^6 + 3x^5 + 6x^4 + 7x^3 + 6x^2 + 3x + 1\)
Similar Questions
Find the middle term in the expansion of \(({2\over 3}x^2 โ {3\over 2x})^{20}\).
Find the 9th term in the expansion of \(({x\over a} โ {3a\over x^2})^{12}\).
Find the 10th term in the binomial expansion of \((2x^2 + {1\over x})^{12}\).
Which is larger \((1.01)^{1000000}\) or 10,000?
Find the middle term in the expansion of \((3x โ {x^3\over 6})^7\).
