Question paper, Set 1

$a(b + c) = ab + ac$

$4a(4a + 5b) = (4a)(4a) + (4a)(5b)$

$= 16a^2 + 20ab$

Answer: $16a^2 + 20ab$

$(a - b)(c - d) = ac - ad - bc + bd$

$(5m - 3n)(5m - 5n) = (5m)(5m) - (5m)(5n) - (3n)(5m) + (3n)(5n)$

$= 25m^2 - 25mn - 15mn + 15n^2$

$= 25m^2 - 40mn + 15n^2$

Answer: $25m^2 - 40mn + 15n^2$

$(a + b)(a - b) = a^2 - b^2$

$(3p - 6q)(3p + 6q) = (3p)^2 - (6q)^2$

$= 9p^2 - 36q^2$

Answer: $9p^2 - 36q^2$

$(a + b)(a - b) = a^2 - b^2$

$(x^5 + 6y^5)(x^5 - 6y^5) = (x^5)^2 - (6y^5)^2$

Using $(ab)^n = a^n \times b^n$: $(x^5)^2 = x^{10}$ and $(6y^5)^2 = 36y^{10}$

$= x^{10} - 36y^{10}$

Answer: $x^{10} - 36y^{10}$

$(a + b)^2 = a^2 + b^2 + 2ab$

$(x + 3y)^2 = (x)^2 + (3y)^2 + 2(x)(3y)$

$= x^2 + 9y^2 + 6xy$

Answer: $x^2 + 9y^2 + 6xy$

$(a + b)^2 = a^2 + b^2 + 2ab$

$\displaystyle \left(5n + \frac{1}{5}\right)^2 = (5n)^2 + \left(\frac{1}{5}\right)^2 + 2(5n)\left(\frac{1}{5}\right)$

$\displaystyle = 25n^2 + \frac{1}{25} + 2n$

Answer: $\displaystyle 25n^2 + \frac{1}{25} + 2n$

$(a - b)^2 = a^2 + b^2 - 2ab$

$(2a - 4b)^2 = (2a)^2 + (4b)^2 - 2(2a)(4b)$

$= 4a^2 + 16b^2 - 16ab$

Answer: $4a^2 + 16b^2 - 16ab$

$(a + b)^2 + (a - b)^2 = 2(a^2 + b^2)$

$(9a + 7b)^2 + (9a - 7b)^2 = 2[(9a)^2 + (7b)^2]$

$= 2(81a^2 + 49b^2)$

$= 162a^2 + 98b^2$

Answer: $2(81a^2 + 49b^2) = 162a^2 + 98b^2$

(i) $(a + b)^2 - (a - b)^2 = 4ab$

$(6x + y)^2 - (6x - y)^2 = 4(6x)(y)$

$(27)^2 - (21)^2 = 24xy$

$729 - 441 = 24xy$

$\displaystyle xy = \frac{288}{24} = 12$

(ii) $(a + b)^2 + (a - b)^2 = 2(a^2 + b^2)$

$(6x + y)^2 + (6x - y)^2 = 2[(6x)^2 + (y)^2]$

$729 + 441 = 2(36x^2 + y^2)$

$\displaystyle 36x^2 + y^2 = \frac{1170}{2} = 585$

Answer: (i) $xy = 12$ (ii) $36x^2 + y^2 = 585$

$(a + b)^2 = (a - b)^2 + 4ab$

$\text{LHS} = (5x - 2y)^2 + 40xy$

$= (5x - 2y)^2 + 4(5x)(2y)$

$= (5x + 2y)^2 = \text{RHS}$

Answer: LHS $= (5x + 2y)^2 =$ RHS

$a^2 + b^2 = (a + b)^2 - 2ab$

(i) $\displaystyle x^2 + \frac{1}{x^2} = (x)^2 + \left(\frac{1}{x}\right)^2$

$\displaystyle = \left(x + \frac{1}{x}\right)^2 - 2 \cdot x \cdot \frac{1}{x}$

$= (3)^2 - 2 = 9 - 2 = 7$

(ii) $\displaystyle x^4 + \frac{1}{x^4} = (x^2)^2 + \left(\frac{1}{x^2}\right)^2$

$\displaystyle = \left(x^2 + \frac{1}{x^2}\right)^2 - 2 \cdot x^2 \cdot \frac{1}{x^2}$

$= (7)^2 - 2 = 49 - 2 = 47$

Answer: (i) $7$ (ii) $47$

$(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$

$(4x + 7y + z)^2 = (4x)^2 + (7y)^2 + (z)^2 + 2(4x)(7y) + 2(7y)(z) + 2(z)(4x)$

$= 16x^2 + 49y^2 + z^2 + 56xy + 14yz + 8zx$

Answer: $16x^2 + 49y^2 + z^2 + 56xy + 14yz + 8zx$

(i) $(a + b - c)^2 = a^2 + b^2 + c^2 + 2ab - 2bc - 2ca$

$(p + 2q - 6r)^2 = (p)^2 + (2q)^2 + (6r)^2 + 2(p)(2q) - 2(2q)(6r) - 2(6r)(p)$

$= p^2 + 4q^2 + 36r^2 + 4pq - 24qr - 12rp$

(ii) $(a - b + c)^2 = a^2 + b^2 + c^2 - 2ab - 2bc + 2ca$

$(7x - 6y + 5z)^2 = (7x)^2 + (6y)^2 + (5z)^2 - 2(7x)(6y) - 2(6y)(5z) + 2(5z)(7x)$

$= 49x^2 + 36y^2 + 25z^2 - 84xy - 60yz + 70zx$

Answer: (i) $p^2 + 4q^2 + 36r^2 + 4pq - 24qr - 12rp$ (ii) $49x^2 + 36y^2 + 25z^2 - 84xy - 60yz + 70zx$

$(a + b)^3 = a^3 + b^3 + 3ab(a + b)$

$(y + 2)^3 = (y)^3 + (2)^3 + 3(y)(2)(y + 2)$

$= y^3 + 8 + 6y(y + 2)$

$= y^3 + 8 + 6y^2 + 12y$

Answer: $y^3 + 8 + 6y^2 + 12y$

$(a - b)^3 = a^3 - b^3 - 3ab(a - b)$

$(3x - 4)^3 = (3x)^3 - (4)^3 - 3(3x)(4)(3x - 4)$

$= 27x^3 - 64 - 36x(3x - 4)$

$= 27x^3 - 64 - 108x^2 + 144x$

Answer: $27x^3 - 64 - 108x^2 + 144x$

(i) $a^2 + b^2 = (a + b)^2 - 2ab$

$p^2 + q^2 = (7)^2 - 2(12) = 49 - 24 = 25$

(ii) $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$

$p^3 + q^3 = (7)^3 - 3(12)(7)$

$= 343 - 252 = 91$

Check with $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$: $(7)(25 - 12) = 7 \times 13 = 91$

Answer: (i) $25$ (ii) $91$

$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

Here $a = 3t$, $b = 5$

$(3t - 5)(9t^2 + 15t + 25) = (3t)^3 - (5)^3 = 27t^3 - 125$

Answer: $27t^3 - 125$

$a^3 - b^3 = (a - b)^3 + 3ab(a - b)$

$a^3 - b^3 = (4)^3 + 3(45)(4)$

$= 64 + 540 = 604$

Answer: $604$

$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$

Here $a = 2p$, $b = 4q$, $c = 1$

$= (2p)^3 + (4q)^3 + (1)^3 - 3(2p)(4q)(1)$

$= 8p^3 + 64q^3 + 1 - 24pq$

Answer: $8p^3 + 64q^3 + 1 - 24pq$

$\text{Take } a = 6,\ b = 9,\ c = -15$

$a + b + c = 6 + 9 - 15 = 0$

If $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$.

$\therefore 6^3 + 9^3 + (-15)^3 = 3(6)(9)(-15) = -2430$

Answer: $-2430$