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Q3. Expand (2x - y)6

Mathematics

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To expand (2x−y)6(2x-y)^6, we use the binomial theorem, which states that (a+b)n=∑k=0n(nk)an−kbk(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k.
In this case, a=2xa = 2x, b=−yb = -y, and n=6n = 6.

  1. Write out the general form of the expansion.

    (2x−y)6=(60)(2x)6(−y)0+(61)(2x)5(−y)1+(62)(2x)4(−y)2+(63)(2x)3(−y)3+(64)(2x)2(−y)4+(65)(2x)1(−y)5+(66)(2x)0(−y)6(2x-y)^6 = \binom{6}{0}(2x)^6(-y)^0 + \binom{6}{1}(2x)^5(-y)^1 + \binom{6}{2}(2x)^4(-y)^2 + \binom{6}{3}(2x)^3(-y)^3 + \binom{6}{4}(2x)^2(-y)^4 + \binom{6}{5}(2x)^1(-y)^5 + \binom{6}{6}(2x)^0(-y)^6

  2. Calculate the binomial coefficients (nk)\binom{n}{k}.

    (60)=1\binom{6}{0} = 1 (61)=6\binom{6}{1} = 6 (62)=6×52×1=15\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15 (63)=6×5×43×2×1=20\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 (64)=(62)=15\binom{6}{4} = \binom{6}{2} = 15 (65)=(61)=6\binom{6}{5} = \binom{6}{1} = 6 (66)=1\binom{6}{6} = 1

  3. Substitute the coefficients and powers into the expansion and simplify each term.

    Term 1: 1⋅(2x)6⋅(−y)0=1⋅64x6⋅1=64x61 \cdot (2x)^6 \cdot (-y)^0 = 1 \cdot 64x^6 \cdot 1 = 64x^6

    Term 2: 6⋅(2x)5⋅(−y)1=6⋅32x5⋅(−y)=−192x5y6 \cdot (2x)^5 \cdot (-y)^1 = 6 \cdot 32x^5 \cdot (-y) = -192x^5y

    Term 3: 15⋅(2x)4⋅(−y)2=15⋅16x4⋅y2=240x4y215 \cdot (2x)^4 \cdot (-y)^2 = 15 \cdot 16x^4 \cdot y^2 = 240x^4y^2

    Term 4: 20⋅(2x)3⋅(−y)3=20⋅8x3⋅(−y3)=−160x3y320 \cdot (2x)^3 \cdot (-y)^3 = 20 \cdot 8x^3 \cdot (-y^3) = -160x^3y^3

    Term 5: 15⋅(2x)2⋅(−y)4=15⋅4x2⋅y4=60x2y415 \cdot (2x)^2 \cdot (-y)^4 = 15 \cdot 4x^2 \cdot y^4 = 60x^2y^4

    Term 6: 6⋅(2x)1⋅(−y)5=6⋅2x⋅(−y5)=−12xy56 \cdot (2x)^1 \cdot (-y)^5 = 6 \cdot 2x \cdot (-y^5) = -12xy^5

    Term 7: 1⋅(2x)0⋅(−y)6=1⋅1⋅y6=y61 \cdot (2x)^0 \cdot (-y)^6 = 1 \cdot 1 \cdot y^6 = y^6

  4. Combine all the terms to get the final expansion.

    (2x−y)6=64x6−192x5y+240x4y2−160x3y3+60x2y4−12xy5+y6(2x-y)^6 = \boxed{64x^6 - 192x^5y + 240x^4y^2 - 160x^3y^3 + 60x^2y^4 - 12xy^5 + y^6}
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