RD Sharma Class 9 Factorisation of Algebraic Expressions Exercise 5.4 Solutions

On this page you will find Maths RD Sharma Class 9 Factorisation of Algebraic Expressions Exercise 5.4 Solutions. The solutions provided here are to help students practice math problems and get better at solving difficult chapter questions. Maths chapter 5 for class 9 deals with the topic of triangles and it is one of the most important chapters.

    Download RD Sharma Class 9 Factorisation of Algebraic Expressions Exercise 5.4 Solutions in PDF

    Students can use these solutions to overcome the fear of maths and the solutions have been designed in such a way that it enables them to discover easy ways to solve different problems. These solutions can help students in refining their maths fluency and problem-solving skills. Students can go through the RD Sharma solutions for class 9 chapter 5 Factorisation of Algebraic Expressions Exercise 5.4 below and it will be beneficial for them.

     

    RD Sharma Class 9 Factorisation of Algebraic Expressions Exercise 5.4 Solutions

    Q1. $a^{3}+8 b^{3}+64 c^{3}-24 a b c$
    SOLUTION :
    $=(a)^{3}+(2 b)^{3}+(4 c)^{3}-3 \times a \times 2 b \times 4 c$
    $=(a+2 b+4 c)\left(a^{2}+(2 b)^{2}+(4 c)^{2}-a \times 2 b-2 b \times 4 c-4 c \times a\right)$
    $\left[\because a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)\right]$
    $=(a+2 b+4 c)\left(a^{2}+4 b^{2}+16 c^{2}-2 a b-8 b c-4 a c\right)$
    $\therefore a^{3}+8 b^{3}+64 c^{3}-24 a b c=(a+2 b+4 c)\left(a^{2}+4 b^{2}+16 c^{2}-2 a b-8 b c-4 a c\right)$

    Q2. $x^{3}-8 y^{3}+27 z^{3}+18 x y z$
    SOLUTION :
    $=x^{3}-(2 y)^{3}+(3 z)^{3}-3 \times x \times(-2 y)(3 z)$
    $=(x+(-2 y)+3 z)\left(x^{2}+(-2 y)^{2}+(3 z)^{2}-x(-2 y)-(-2 y)(3 z)-3 z(x)\right)$
    $\left[\because a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)\right]$
    $=(x+(-2 y)+3 z)\left(x^{2}+4 y^{2}+9 z^{2}+2 x y+6 y z-3 z x\right)$
    $\therefore x^{3}-8 y^{3}+27 z^{3}+18 x y z=(x+(-2 y)+3 z)\left(x^{2}+4 y^{2}+9 z^{2}+2 x y+6 y z-3 z x\right)$

    Q3. $\frac{1}{27} x^{3}-y^{3}+125 z^{3}+5 x y z$
    SOLUTION :
    $=\left(\frac{x}{3}\right)^{3}+(-y)^{3}+(5 z)^{3}-3 \times \frac{x}{3}(-y)(5 z)$
    $=\left(\frac{x}{3}+(-y)+5 z\right)\left(\left(\frac{x}{3}\right)^{2}+(-y)^{2}+(5 z)^{2}-\frac{x}{3}(-y)-(-y) 5 z-5 z\left(\frac{x}{3}\right)\right)$
    $=\left(\frac{x}{3}-y+5 z\right)\left(\frac{x^{2}}{9}+y^{2}+25 z^{2}+\frac{x y}{3}+5 y z-\frac{5 z x}{3}\right)$
    $\therefore \frac{1}{27} x^{3}-y^{3}+125 z^{3}+5 x y z=\left(\frac{x}{3}-y+5 z\right)\left(\frac{x^{2}}{9}+y^{2}+25 z^{2}+\frac{x y}{3}+5 y z-\frac{5 z x}{3}\right)$

    Q4. $8 x^{3}+27 y^{3}-216 z^{3}+108 x y z$
    SOLUTION :
    $=(2 x)^{3}+(3 y)^{3}+(-6 y)^{3}-3(2 x)(3 y)(-6 z)$
    $=(2 x+3 y+(-6 z))\left((2 x)^{2}+(3 y)^{2}\right.\left.+(-6 z)^{2}-2 x \times 3 y-3 y(-6 z)-(-6 z) 2 x\right)$
    $=(2 x+3 y+(-6 z))\left(4 x^{2}+9 y^{2}+36 z^{2}-6 x y+18 y z+12 z x\right)$
    $\therefore 8 x^{3}+27 y^{3}-216 z^{3}+108 x y z=(2 x+3 y+(-6 z))\left(4 x^{2}+9 y^{2}+36 z^{2}-6 x y+18 y z+12 z x\right)$

    Q5. $125+8 x^{3}-27 y^{3}+90 x y$
    SOLUTION :
    $=(5)^{3}+(2 x)^{3}+(-3 y)^{3}-3 \times 5 \times 2 x \times(-3 y)$
    $=(5+2 x+(-3 y))\left(5^{2}+(2 x)^{2}+(-3 y)^{2}-5(2 x)-2 x(-3 y)-(-3 y) 5\right)$
    $=(5+2 x-3 y)\left(25+4 x^{2}+9 y^{2}-10 x+6 x y+15 y\right)$
    $\therefore 125+8 x^{3}-27 y^{3}+90 x y=(5+2 x-3 y)\left(25+4 x^{2}+9 y^{2}-10 x+6 x y+15 y\right)$

    Q6. $(3 x-2 y)^{3}+(2 y-4 z)^{3}+(4 z-3 x)^{3}$
    SOLUTION :
    Let $(3 x-2 y)=a,(2 y-4 z)=b,(4 z-3 x)=c$
    $\therefore \mathrm{a}+\mathrm{b}+\mathrm{c}=3 \mathrm{x}-2 \mathrm{y}+2 \mathrm{y}-4 \mathrm{z}+4 \mathrm{z}-3 \mathrm{x}=0$
    $\because \mathrm{a}+\mathrm{b}+\mathrm{c}=0 \therefore \mathrm{a}^{3}+\mathrm{b}^{3}+\mathrm{c}^{3}=3 \mathrm{abc}$
    $=3(3 x-2 y)(2 y-4 z)(4 z-3 x)$
    $\therefore(3 x-2 y)^{3}+(2 y-4 z)^{3}+(4 z-3 x)^{3}=3(3 x-2 y)(2 y-4 z)(4 z-3 x)$

    Q7. $(2 x-3 y)^{3}+(4 z-2 x)^{3}+(3 y-4 z)^{3}$
    SOLUTION :
    Let $2 x-3 y=a, 4 z-2 x=b, 3 y-4 z=c$
    $\therefore \mathrm{a}+\mathrm{b}+\mathrm{c}=2 \mathrm{x}-3 \mathrm{y}+4 \mathrm{z}-2 \mathrm{x}+3 \mathrm{y}-4 \mathrm{z}=0$
    $\because \mathrm{a}+\mathrm{b}+\mathrm{c}=0 \therefore \mathrm{a}^{3}+\mathrm{b}^{3}+\mathrm{c}^{3}=3 \mathrm{abc}$
    $=3(2 x-3 y)(4 z-2 x)(3 y-4 z)$
    $\therefore(2 x-3 y)^{3}+(4 z-2 x)^{3}+(3 y-4 z)^{3}=3(2 x-3 y)(4 z-2 x)(3 y-4 z)$

    Q8. $\left[\frac{x}{2}+y+\frac{z}{3}\right]^{3}+\left[\frac{x}{3}-\frac{2 y}{3}+z\right]^{3}+\left[-\frac{5 x}{6}-\frac{y}{3}-\frac{4 z}{3}\right]^{3}$
    SOLUTION :
    Let $\left[\frac{x}{2}+y+\frac{z}{3}\right]=a,\left[\frac{x}{3}-\frac{2 y}{3}+z\right]=b,\left[-\frac{5 x}{6}-\frac{y}{3}-\frac{4 z}{3}\right]=c$
    $\mathrm{a}+\mathrm{b}+\mathrm{c}=\frac{\mathrm{x}}{2}+\mathrm{y}+\frac{\mathrm{z}}{3}+\frac{\mathrm{x}}{3}-\frac{2 \mathrm{y}}{3}+\mathrm{z}-\frac{5 \mathrm{x}}{6}-\frac{\mathrm{y}}{3}-\frac{4 \mathrm{z}}{3}$
    $\mathrm{a}+\mathrm{b}+\mathrm{c}=\left(\frac{\mathrm{x}}{2}+\frac{\mathrm{x}}{3}-\frac{5 \mathrm{x}}{6}\right)+\left(\mathrm{y}-\frac{2 \mathrm{y}}{3}-\frac{\mathrm{y}}{3}\right)+\left(\frac{\mathrm{z}}{3}+\mathrm{z}-\frac{4 \mathrm{z}}{3}\right)$
    $\mathrm{a}+\mathrm{b}+\mathrm{c}=\frac{3 \mathrm{x}}{6}+\frac{2 \mathrm{x}}{6}-\frac{5 \mathrm{x}}{6}+\frac{3 \mathrm{y}}{3}-\frac{2 \mathrm{y}}{3}-\frac{\mathrm{y}}{3}+\frac{\mathrm{z}}{3}+\frac{3 \mathrm{z}}{3}-\frac{4 \mathrm{z}}{3}$
    $a+b+c=\frac{5 x-5 x}{6}+\frac{3 y-3 y}{3}+\frac{4 z-4 z}{3}$
    $a+b+c=0$
    $\because a+b+c=0$ $\therefore \mathrm{a}^{3}+\mathrm{b}^{3}+\mathrm{c}^{3}=3 \mathrm{abc}$
    $=3\left(\frac{x}{2}+y+\frac{z}{3}\right)\left(\frac{x}{3}-\frac{2 y}{3}+z\right)\left(-\frac{5 x}{6}-\frac{y}{3}-\frac{4 z}{3}\right)$
    $\therefore\left[\frac{x}{2}+y+\frac{z}{3}\right]^{3}+\left[\frac{x}{3}-\frac{2 y}{3}+z\right]^{3}+\left[-\frac{5 x}{6}-\frac{y}{3}-\frac{4 z}{3}\right]^{3} =3\left(\frac{x}{2}+y+\frac{z}{3}\right)\left(\frac{x}{3}-\frac{2 y}{3}+z\right)\left(-\frac{5 x}{6}-\frac{y}{3}-\frac{4 z}{3}\right)$

    Q9. $(a-3 b)^{3}+(3 b-c)^{3}+(c-a)^{3}$
    SOLUTION :
    Let $a-3 b=x, 3 b-c=y, c-a=z$
    $x+y+z=a-3 b+3 b-c+c-a=0$
    $(\because \mathrm{x}+\mathrm{y}+\mathrm{z}=0)$ $\therefore \mathrm{x}^{3}+\mathrm{y}^{3}+\mathrm{z}^{3}=3 \mathrm{xyz}$
    $=3(a-3 b)(3 b-c)(c-a)$
    $\therefore(a-3 b)^{3}+(3 b-c)^{3}+(c-a)^{3}=3(a-3 b)(3 b-c)(c-a)$

    Q 10 . $2 \sqrt{2} a^{3}+3 \sqrt{3} b^{3}+c^{3}-3 \sqrt{6 a b c}$
    SOLUTION :
    $=(\sqrt{2} a)^{3}+(\sqrt{3} b)^{3}+c^{3}-3 \times \sqrt{2 a} \times \sqrt{3 b} \times c$
    $=(\sqrt{2 a}+\sqrt{3 b}+c)\left((\sqrt{2 a})^{2}+(\sqrt{3 b})^{2}+c^{2}-(\sqrt{2 a})(\sqrt{3 b})-(\sqrt{3 b}) c-(\sqrt{2 a} c)\right)$
    $=(\sqrt{2 a}+\sqrt{3 b}+c)\left(2 a^{2}+3 b^{2}+c^{2}-\sqrt{6} a b-\sqrt{3} b c-\sqrt{2} a c\right)$
    $\therefore 2 \sqrt{2} a^{3}+3 \sqrt{3} b^{3}+c^{3}-3 \sqrt{6 a b c}=(\sqrt{2 a}+\sqrt{3 b}+c)\left(2 a^{2}+3 b^{2}+c^{2}-\sqrt{6} a b-\sqrt{3} b c-\sqrt{2} a c\right)$

    Q11. $3 \sqrt{3} a^{3}-b^{3}-5 \sqrt{5} c^{3}-3 \sqrt{15} a b c$
    SOLUTION :
    $=(\sqrt{3} a)^{3}+(-b)^{3}-(\sqrt{5} c)^{3}-3(\sqrt{3 a})(-b)(-\sqrt{5} c)$
    $=(\sqrt{3} a+(-b)+(-\sqrt{5} c))\left((\sqrt{3} a)^{2}+(-b)^{2}+(-\sqrt{5} c)^{2}-\sqrt{3} a(-b)-(-b)(-\sqrt{5} c)\right.-(-\sqrt{5} c) \sqrt{3} a)$
    $=(\sqrt{3} a-b-\sqrt{5} c)\left(3 a^{2}+b^{2}+5 c^{2}+\sqrt{3} a b-\sqrt{5} b c+\sqrt{15} a c\right)$
    $\therefore 3 \sqrt{3} \mathrm{a}^{3}-\mathrm{b}^{3}-5 \sqrt{5} \mathrm{c}^{3}-3 \sqrt{15} \mathrm{abc}=(\sqrt{3} \mathrm{a}-\mathrm{b}-\sqrt{5} \mathrm{c})\left(3 \mathrm{a}^{2}+\mathrm{b}^{2}+5 \mathrm{c}^{2}+\sqrt{3} \mathrm{ab}-\sqrt{5} \mathrm{bc}+\sqrt{15} \mathrm{ac}\right)$

    Q 12 . $8 x^{3}-125 y^{3}+216+180 x y$
    SOLUTION :
    $=(2 x)^{3}+(-5 y)^{3}+6^{3}-3 \times(2 x)(-5 y)(6)$
    $=(2 x+(-5 y)+6)\left((2 x)^{2}+(-5 y)^{2}+6^{2}-2 x \times(-5 y)-(-5 y) 6-6(2 x)\right)$
    $=(2 x-5 y+6)\left(4 x^{2}+25 y^{2}+36+10 x y+30 y-12 x\right)$
    $\therefore 8 x^{3}-125 y^{3}+216+180 x y=(2 x-5 y+6)\left(4 x^{2}+25 y^{2}+36+10 x y+30 y-12 x\right)$

    Q13 $\cdot 2 \sqrt{2} \mathrm{a}^{3}+16 \sqrt{2} \mathrm{~b}^{3}+\mathrm{c}^{3}-12 \mathrm{abc}$
    SOLUTION :
    $=(\sqrt{2} a)^{3}+(2 \sqrt{2} b)^{3}+c^{3}-3 \times \sqrt{2} a \times 2 \sqrt{2} b \times c$
    $=(\sqrt{2} a+2 \sqrt{2} b+c)\left((\sqrt{2} a)^{2}+(2 \sqrt{2} b)^{2}+c^{2}-(\sqrt{2} a)(2 \sqrt{2} b)-(2 \sqrt{2} b) c-(\sqrt{2} a) c\right)$
    $=(\sqrt{2} a+2 \sqrt{2} b+c)\left(2 a^{2}+8 b^{2}+c^{2}-4 a b-2 \sqrt{2} b c-\sqrt{2} a c\right)$
    $\therefore 2 \sqrt{2} \mathrm{a}^{3}+16 \sqrt{2} \mathrm{~b}^{3}+\mathrm{c}^{3}-12 \mathrm{abc}=(\sqrt{2} \mathrm{a}+2 \sqrt{2} \mathrm{~b}+\mathrm{c})\left(2 \mathrm{a}^{2}+8 \mathrm{~b}^{2}+\mathrm{c}^{2}-4 \mathrm{ab}-2 \sqrt{2} \mathrm{bc}-\sqrt{2} \mathrm{ac}\right)$

    Q 14 . Find the value of $\mathrm{x}^{3}+\mathrm{y}^{3}-12 \mathrm{xy}+64$ when $x+y=-4$.
    SOLUTION :
    $=x^{3}+y^{3}+64-12 x y$
    $=x^{3}+y^{3}+4^{3}-3(x)(y)(4)$
    $=(x+y+4)\left(x^{2}+y^{2}+4^{2}-x y-y \times 4-4 \times x\right)$
    $=(-4+4)\left(x^{2}+y^{2}+16-x y-4 y-4 x\right)$ $[\because x+y=-4]$
    $=0$
    $\therefore x^{3}+y^{3}-12 x y+64=0$

    Q 15 . MULTIPLY :
    (i) $\cdot x^{2}+y^{2}+z^{2}-x y+x z+y z$ by $x+y-z$
    SOLUTION :
    $=\left(x^{2}+y^{2}+z^{2}-x y+x z+y z\right)(x+y-z)$
    $=x^{3}+y^{3}+z^{3}-3 x y z$
    (ii) $x^{2}+4 y^{2}+z^{2}+2 x y+x z-2 y z$ by $x-2 y-z$
    SOLUTION :
    $x^{2}+(-2 y)^{2}+(-z)^{2}-(-2 y)(-z)-(-z)(x)=x^{3}+(-2 y)^{3}+(-z)^{3}-3 x(-2 y)(-z)$
    $\Rightarrow x^{2}+4 y^{2}+z^{2}+2 x y-2 y z+z x=x^{3}-8 y^{3}-z^{3}-6 x y z$
    (iii) $x^{2}+4 y^{2}+2 x y-3 x+6 y+9$ by $(x-2 y+3)$
    SOLUTION :
    $(x)^{2}+(-2 y)^{2}+(3)^{2}-(x)(-2 y)-(-2 y)(3)-3(x)=(x)^{3}+(-2 y)^{3}+3^{3}-3(x)(-2 y)(3)$
    $\Rightarrow x^{2}+4 y^{2}+9+2 x y+6 y-3 x=x^{3}-8 y^{3}+27+18 x y$
    (iv) $.9 x^{2}+25 y^{2}+15 x y+12 x-20 y+16$ by $3 x-5 y+4$
    SOLUTION :
    $(3 x)^{2}+(5 y)^{2}+4^{2}-(-3 x)(5 y)-(5 y)(4)-(4)(-3 x)=(-3 x)^{3}+(5 y)^{3}+4^{3}$
    $-3(-3 x)(5 y)(4)$
    $\Rightarrow 9 x^{2}+25 y^{2}+16+15 x y-20 y+12 x=-27 x^{3}+125 y^{3}+64+180 x y$

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