3 Mark Question

Question 13 Marks

Factorise: $1+\frac{\mathrm{q}^3}{125}$

Answer

$\begin{aligned} & \text { } 1+\frac{\mathrm{q}^3}{125} \\ & =1^3+\left(\frac{\mathrm{q}}{5}\right)^3 \\ & \text { Here, } a=1 \text { and } b=\frac{q}{5} \\ & \therefore \quad 1+\frac{\mathrm{q}^3}{125}=\left(1+\frac{\mathrm{q}}{5}\right)\left[1^2-1\left(\frac{\mathrm{q}}{5}\right)+\left(\frac{\mathrm{q}}{5}\right)^2\right] \\ & \ldots\left[\because a^3+b^3=(a+b)\left(a^2-a b+b^2\right)\right] \\ & =\left(1+\frac{q}{5}\right)\left(1-\frac{q}{5}+\frac{q^2}{25}\right) \\ & \end{aligned}$

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Question 23 Marks

Factorise: $a^3+\frac{8}{a^3}$

Answer

$\begin{aligned} & \text { } a^3+\frac{8}{a^3} \\ & =a^3+\left(\frac{2}{a}\right)^3 \\ & \text { Here, } A=a \text { and } B=\frac{2}{a} \\ & \therefore \quad a^3+\frac{8}{a^3}=\left(a+\frac{2}{a}\right)\left[a^2-a\left(\frac{2}{a}\right)+\left(\frac{2}{a}\right)^2\right] \\ & \ldots\left[\because A^3+B^3=(A+B)\left(A^2-A B+B^2\right)\right] \\ & =\left(a+\frac{2}{a}\right)\left(a^2-2+\frac{4}{a^2}\right) \\ & \end{aligned}$

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Question 33 Marks

Factorise: $y^3+\frac{1}{8 y^3}$

Answer

\text { } y^3+\frac{1}{8 y^3}

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Question 43 Marks

Factorise: $24 a^3+81 b^3$

Answer

24a³ + 81b³
…[Taking out the common factor 3]
= 3 [(2a)³ + (3b)³]
Here, A = 2a and B = 3b
∴ 24a³ + 81b³
= 3 {(2a + 3b) [(2a)² – (2a)(3b) + (3b)²]}
…[∵ A³ + B³ = (A + B) (A² – AB + B²)]
= 3(2a + 3b)(4a² – 6ab + 9b²)

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Question 53 Marks

Factorise: $2 l^3+432 m^3$

Answer

2l³ + 432m³
= 2 (l³ + 216m³)
… [Taking out the common factor 2]
= 2[l³ + (6m)³]
Here, a = l and b = 6m
2l³ + 432m³ = 2 {(l + 6m)[l² – l(6m) + (6m)²]}
…[∵ a³ + b³ = (a + b)(a² – ab + b²)]
= 2(l + 6m)(l² – 6lm + 36m²)

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Question 63 Marks

Factorise: $125 \mathrm{k}^3+27 \mathrm{~m}^3$

Answer

125k³ + 27m³
= (5k)³ + (3m)³
Here, a = 5k and b = 3m
∴ 125k³ + 27m³
= (5k + 3m) [(5k)² – (5k)(3m) + (3m)²]
…[∵ a³ + b³ = (a + b)(a² – ab + b²)]
= (5k + 3m)(25k² – 15km + 9m²)

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Question 73 Marks

Factorise: $125 p^3+q^3$

Answer

125p³ + q³
= (5p)³ + q³
Here, a = 5p and b = q
∴ 125p³ + q³ = (5p + q)[(5p)² – (5p)(q) + q²]
…[∵ a³ + b³ = (a + b)(a² – ab + b²)]
= (5p + q)(25p² – 5pq + q²)

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Question 83 Marks

Factorise: $\mathrm{x}^3+64 \mathrm{y}^3$

Answer

x³ + 64y³
= x³ + (4y)³
Here, a = x and b = 4y
∴ x³ + 64y³ = (x + 4y) [x² – x(4y) + (4y)²]
….[∵ a³ + b³ = (a + b)(a² – ab + b²)]
= (x + 4y)(x² – 4xy + 16y²)

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