Answer :

Let us understand what a G.C.D, Greatest Common Divisor and L.C.M, Least Common Multiple is.

The greatest common divisor of two or more integers, which are not all zero, is the largest positive integer that divides each of the integers.

Least common multiple of two or more integers, is the smallest positive integer that is divisible by these two or more integers.

To find G.C.D of (y^{3} – 8), (y^{3} – 4y^{2} + 4y) and (y^{2} + y – 6), let us write down factors of each term.

For factorization of y^{3} – 8,

Factorization of y^{3} – 8 = Factorization of y^{3} – 2^{3}

[∵, y^{3} – 8 = y^{3} – 2^{3}, as 2^{3} = 2 × 2 × 2 = 8]

⇒ y^{3} – 8 = (y – 2)(y^{2} + 2^{2} + (2)(y))

⇒ y^{3} – 8 = (y – 2)(y^{2} + 4 + 2y)

⇒ y^{3} – 8 = (y – 2)(y^{2} + 2y + 4)

[∵, by algebraic identity, a^{3} – b^{3} = (a – b)(a^{2} + b^{2} + ab)]

For factorization of y^{3} – 4y^{2} + 4y,

⇒ y^{3} – 4y^{2} + 4y = y(y^{2} – 4y + 4)

⇒ y^{3} – 4y^{2} + 4y = y(y^{2} – (2y + 2y) + 4) [∵, Sum of -2 and -2 is -4 and multiplication is 4]

⇒ y^{3} – 4y^{2} + 4y = y(y^{2} – 2y – 2y + 4)

⇒ y^{3} – 4y^{2} + 4y = y(y(y – 2) – 2(y – 2)) [∵, common from the first two terms is y and last two terms is -2]

⇒ y^{3} – 4y^{2} + 4y = y((y – 2)(y – 2)) [∵, common from the two terms is (y – 2)]

⇒ y^{3} – 4y^{2} + 4y = y(y – 2)(y – 2)

⇒ y^{3} – 4y^{2} + 4y = y(y – 2)^{2}

For factorization of y^{2} + y – 6,

⇒ y^{2} + y – 6 = y^{2} + 3y – 2y – 6 [∵, Sum of 3 and -2 is 1 and multiplication is -6]

⇒ y^{2} + y – 6 = y(y + 3) – 2(y + 3) [∵, common from the first two terms is y and last two terms is -2]

⇒ y^{2} + y – 6 = (y + 3)(y – 2) [∵, common from the two terms is (y + 3)]

Find the factors that these two lists share in common.

Factors of y^{3} – 8 = (y – 2), (y^{2} + 2y + 4)

Factors of y^{3} – 4y^{2} + 4y = y, (y – 2), (y – 2)

Factors of y^{2} + y – 6 = (y + 3), (y – 2)

Common factor that is found in these three terms is (y – 2).

So, gcd is (y – 2).

To find L.C.D of (y^{3} – 8), (y^{3} – 4y^{2} + 4y) and (y^{2} + y – 6), let us write down factors of each term.

Then, we need to find the factor of highest power.

Factors of y^{3} – 8 = (y – 2), (y^{2} + 2y + 4)

Factors of y^{3} – 4y^{2} + 4y = y, (y – 2)^{2}

Factors of y^{2} + y – 6 = (y + 3), (y – 2)

The factors are y, (y – 2)^{2}, (y + 3) and (y^{2} + 2y + 4).

Multiplying these factors, we get

y × (y – 2)^{2} × (y + 3) × (y^{2} + 2y + 4) = y(y – 2)^{2}(y + 3)(y^{2} + 2y + 4)

So, lcm is y(y – 2)^{2}(y + 3)(y^{2} + 2y + 4).

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Let's find G.C.D.West Bengal - Mathematics

Let's find the G.West Bengal - Mathematics

Let's find the G.West Bengal - Mathematics

Let’s find G.C.D West Bengal - Mathematics

Let’s find G.C.D West Bengal - Mathematics

Let’s find G.C.D West Bengal - Mathematics

Let’s find G.C.D West Bengal - Mathematics

Let’s find G.C.D West Bengal - Mathematics

Let’s find G.C.D West Bengal - Mathematics

Let’s find G.C.D West Bengal - Mathematics