Maharashtra Board Class 8 Maths Chapter 6 Factorisation of Algebraic Expressions Set 6.2 Solutions

Question 1. Factorise:
(i) x³ + 64y³
(ii) 125p³ + q³
(iii) 125k³ + 27m³
(iv) 2l³ + 432m³
(v) 24a³ + 81b³
(vi) y³ + \(\frac{1}{8y^3}\)
(vii) a³ + \(\frac{8}{a^3}\)
(viii) 1 + \(\frac{q^3}{125}\)
Answer:
(i) x³ + 64y³
= \(x^3 + (4y)^3\)
Here, a = \(x\) and b = \(4y\)
\(\therefore x^3 + 64y^3 = (x + 4y) [x^2 - x(4y) + (4y)^2]\)
\[\therefore a^3 + b^3 = (a + b)(a^2 - ab + b^2)\]
= \((x + 4y)(x^2 - 4xy + 16y^2)\)

(ii) 125p³ + q³
= \((5p)^3 + q^3\)
Here, a = \(5p\) and b = \(q\)
\(\therefore 125p^3 + q^3 = (5p + q)[(5p)^2 - (5p)(q) + q^2]\)
\[\therefore a^3 + b^3 = (a + b)(a^2 - ab + b^2)\]
= \((5p + q)(25p^2 - 5pq + q^2)\)

(iii) 125k³ + 27m³
= \((5k)^3 + (3m)^3\)
Here, a = \(5k\) and b = \(3m\)
\(\therefore 125k^3 + 27m^3\)
= \((5k + 3m) [(5k)^2 - (5k)(3m) + (3m)^2]\)
\[\therefore a^3 + b^3 = (a + b)(a^2 - ab + b^2)\]
= \((5k + 3m)(25k^2 - 15km + 9m^2)\)

(iv) 2l³ + 432m³
= \(2 (l^3 + 216m^3)\)
[Taking out the common factor 2]
= \(2[l^3 + (6m)^3]\)
Here, a = \(l\) and b = \(6m\)
\(2l^3 + 432m^3 = 2 \{(l + 6m)[l^2 - l(6m) + (6m)^2]\}\)
\[\therefore a^3 + b^3 = (a + b)(a^2 - ab + b^2)\]
= \(2(l + 6m)(l^2 - 6lm + 36m^2)\)

(v) 24a³ + 81b³
[Taking out the common factor 3]
= \(3 [(2a)^3 + (3b)^3]\)
Here, A = \(2a\) and B = \(3b\)
\(\therefore 24a^3 + 81b^3\)
= \(3 \{(2a + 3b) [(2a)^2 - (2a)(3b) + (3b)^2]\}\)
\[\therefore A^3 + B^3 = (A + B) (A^2 - AB + B^2)\]
= \(3(2a + 3b)(4a^2 - 6ab + 9b^2)\)

(vi) y³ + \(\frac{1}{8y^3}\)
= \(y^3 + \left(\frac{1}{2y}\right)^3\)
Here, a = \(y\) and b = \(\frac{1}{2y}\)
\(\therefore y^3 + \frac{1}{8y^3} = \left(y + \frac{1}{2y}\right)\left(y^2 - y\left(\frac{1}{2y}\right) + \left(\frac{1}{2y}\right)^2\right)\)
\[\therefore a^3+b^3 = (a + b)(a^2 - ab + b^2)\]
= \(\left(y + \frac{1}{2y}\right)\left(y^2 - \frac{1}{2} + \frac{1}{4y^2}\right)\)

(vii) a³ + \(\frac{8}{a^3}\)
= \(a^3 + \left(\frac{2}{a}\right)^3\)
Here, A = \(a\) and B = \(\frac{2}{a}\)
\(\therefore a^3 + \frac{8}{a^3} = \left(a + \frac{2}{a}\right)\left(a^2 - a\left(\frac{2}{a}\right) + \left(\frac{2}{a}\right)^2\right)\)
\[\therefore A^3 + B^3 = (A+B) (A^2 - AB + B^2)\]
= \(\left(a + \frac{2}{a}\right)\left(a^2 - 2 + \frac{4}{a^2}\right)\)

(viii) 1 + \(\frac{q^3}{125}\)
= \(1^3 + \left(\frac{q}{5}\right)^3\)
Here, a = \(1\) and b = \(\frac{q}{5}\)
\(\therefore 1 + \frac{q^3}{125} = \left(1 + \frac{q}{5}\right)\left[1^2 - 1\left(\frac{q}{5}\right) + \left(\frac{q}{5}\right)^2\right]\)
\[\therefore a^3+b^3 = (a + b)(a^2 - ab + b^2)\]
= \(\left(1 + \frac{q}{5}\right)\left(1 - \frac{q}{5} + \frac{q^2}{25}\right)\)
In simple words: This question involves factorizing various sums of cubes. The key is to recognize the pattern \(a^3 + b^3 = (a+b)(a^2-ab+b^2)\), identify the 'a' and 'b' terms for each expression, and then substitute them into the formula. For some parts, a common factor needs to be extracted first.

🎯 Exam Tip: Remember to identify the cubic terms correctly and apply the factorization formula for the sum of cubes without errors. Pay attention to common factors, especially when dealing with coefficients other than 1.