Without actually calculating the cubes, find the value of each of the following:
(i) (–12)3 + (7)3 + (5)3
(ii) (28)3 + (–15)3 + (–13)3
(i) (–12)3 + (7)3 + (5)3
Let x = -12, y = 7, and z = 5
It can be observed that, x + y + z = -12 + 7 + 5 = 0
It is known that if x + y + z = 0, then
x3 + y3 + z3 = 3xyz
∴ (-12)3 + (7)3 + (5)3 = 3(-12) (7) (5) = -1260
(ii) (28)3 + (–15)3 + (–13)3
Let x = 28, y = -15, and z = -13
It can be observed that,
x + y + z = 28 + (-15) + (-13) = 28 - 28 = 0
It is known that if x + y + z = 0,
then x3 + y3 + z3 = 3xyz
∴ (28)3 + (-15)3 + (-13)3 = 3(28) (-15) (-13)=16380
Write the degree of each of the following polynomials:
(i) 5x 3 + 4x 2 + 7x (ii) 4 – y 2 (iii) 5t – √7 (iv) 3.
Classify the following as linear, quadratic and cubic polynomials:
(i) x 2 + x (ii) x – x 3 (iii) y + y 2 + 4 (iv) 1 + x (v) 3t (vi) r 2 (vii) 7x 3
Find the zero of the polynomial in each of the following cases:
(i) p(x) = x + 5 (ii) p(x) = x – 5 (iii) p(x) = 2x + 5 (iv) p(x) = 3x – 2 (v) p(x) = 3x
(vi) p(x) = ax, a ≠ 0 (vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.
(i) The kind of person the doctor is (money, possessions)
(ii) The kind of person he wants to be (appearance, ambition)
ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see Fig. 7.17). Prove that
(i) ∆ ABD ≅ ∆ BAC
(ii) BD = AC
(iii) ∠ ABD = ∠ BAC.
