Question paper, Set 5

$(a - b)c = ac - bc$

$(6p - 3q) \times 9p = (6p)(9p) - (3q)(9p)$

$= 54p^2 - 27pq$

Answer: $54p^2 - 27pq$

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

$(4p - 2q)(2p + 7q) = (4p)(2p) + (4p)(7q) - (2q)(2p) - (2q)(7q)$

$= 8p^2 + 28pq - 4pq - 14q^2$

$= 8p^2 + 24pq - 14q^2$

Answer: $8p^2 + 24pq - 14q^2$

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

$(7x - y)(7x + y) = (7x)^2 - (y)^2$

$= 49x^2 - y^2$

Answer: $49x^2 - y^2$

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

$(5a^4 + 5b^2)(5a^4 - 5b^2) = (5a^4)^2 - (5b^2)^2$

Using $(ab)^n = a^n \times b^n$: $(5a^4)^2 = 25a^8$ and $(5b^2)^2 = 25b^4$

$= 25a^8 - 25b^4$

Answer: $25a^8 - 25b^4$

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

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

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

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

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

$\displaystyle \left(5y + \frac{1}{9}\right)^2 = (5y)^2 + \left(\frac{1}{9}\right)^2 + 2(5y)\left(\frac{1}{9}\right)$

$\displaystyle = 25y^2 + \frac{1}{81} + \frac{10y}{9}$

Answer: $\displaystyle 25y^2 + \frac{1}{81} + \frac{10y}{9}$

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

$(6p - 9q)^2 = (6p)^2 + (9q)^2 - 2(6p)(9q)$

$= 36p^2 + 81q^2 - 108pq$

Answer: $36p^2 + 81q^2 - 108pq$

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

$(2p + 4q)^2 + (2p - 4q)^2 = 2[(2p)^2 + (4q)^2]$

$= 2(4p^2 + 16q^2)$

$= 8p^2 + 32q^2$

Answer: $2(4p^2 + 16q^2) = 8p^2 + 32q^2$

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

$(3x + 4y)^2 - (3x - 4y)^2 = 4(3x)(4y)$

$(28)^2 - (20)^2 = 48xy$

$784 - 400 = 48xy$

$\displaystyle xy = \frac{384}{48} = 8$

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

$(3x + 4y)^2 + (3x - 4y)^2 = 2[(3x)^2 + (4y)^2]$

$784 + 400 = 2(9x^2 + 16y^2)$

$\displaystyle 9x^2 + 16y^2 = \frac{1184}{2} = 592$

Answer: (i) $xy = 8$ (ii) $9x^2 + 16y^2 = 592$

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

Consider $(8p + 6q)^2 - 192pq$

$= (8p + 6q)^2 - 4(8p)(6q)$

$= (8p - 6q)^2$

Answer: $(8p - 6q)^2$

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

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

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

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

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

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

$= 47 - 2 = 45$

Answer: (i) $47$ (ii) $45$

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

$(7a + 2b + 6c)^2 = (7a)^2 + (2b)^2 + (6c)^2 + 2(7a)(2b) + 2(2b)(6c) + 2(6c)(7a)$

$= 49a^2 + 4b^2 + 36c^2 + 28ab + 24bc + 84ca$

Answer: $49a^2 + 4b^2 + 36c^2 + 28ab + 24bc + 84ca$

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

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

$= 9p^2 + q^2 + 36r^2 - 6pq - 12qr + 36rp$

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

$(4p + 7q - 2r)^2 = (4p)^2 + (7q)^2 + (2r)^2 + 2(4p)(7q) - 2(7q)(2r) - 2(2r)(4p)$

$= 16p^2 + 49q^2 + 4r^2 + 56pq - 28qr - 16rp$

Answer: (i) $9p^2 + q^2 + 36r^2 - 6pq - 12qr + 36rp$ (ii) $16p^2 + 49q^2 + 4r^2 + 56pq - 28qr - 16rp$

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

$(4y + 1)^3 = (4y)^3 + (1)^3 + 3(4y)(1)(4y + 1)$

$= 64y^3 + 1 + 12y(4y + 1)$

$= 64y^3 + 1 + 48y^2 + 12y$

Answer: $64y^3 + 1 + 48y^2 + 12y$

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

$(4x - 1)^3 = (4x)^3 - (1)^3 - 3(4x)(1)(4x - 1)$

$= 64x^3 - 1 - 12x(4x - 1)$

$= 64x^3 - 1 - 48x^2 + 12x$

Answer: $64x^3 - 1 - 48x^2 + 12x$

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

$m^2 + n^2 = (15)^2 - 2(56) = 225 - 112 = 113$

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

$m^3 + n^3 = (15)^3 - 3(56)(15)$

$= 3375 - 2520 = 855$

Check with $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$: $(15)(113 - 56) = 15 \times 57 = 855$

Answer: (i) $113$ (ii) $855$

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

Here $a = 2x$, $b = 3$

$(2x - 3)(4x^2 + 6x + 9) = (2x)^3 - (3)^3 = 8x^3 - 27$

Answer: $8x^3 - 27$

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

$p^3 - q^3 = (4)^3 + 3(60)(4)$

$= 64 + 720 = 784$

Answer: $784$

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

Here $a = 2m$, $b = 4n$, $c = 5$

$= (2m)^3 + (4n)^3 + (5)^3 - 3(2m)(4n)(5)$

$= 8m^3 + 64n^3 + 125 - 120mn$

Answer: $8m^3 + 64n^3 + 125 - 120mn$

$\text{Take } a = 12,\ b = -7,\ c = -5$

$a + b + c = 12 - 7 - 5 = 0$

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

$\therefore 12^3 - 7^3 - 5^3 = 3(12)(-7)(-5) = 1260$

Answer: $1260$