Q. 205.0( 5 Votes )

# Simplify:

Answer :

(i) 2x^{2}(x^{3} – x) – 3x(x^{4} + 2x) – 2(x^{4} – 3x^{2})

= 2x^{5} – 2x^{3} – 3x^{5} – 6x^{2} – 2x^{4} + 6x^{2}

= -x^{5} – 2x^{4} – 2x^{3}

(ii) x^{3}y(x^{2} – 2x) + 2xy(x^{3} – x^{4})

= x^{5}y – 2x^{4}y + 2x^{4}y – 2x^{5}y

= -x^{5}y

(iii) 3a^{2} + (a + 2) – 3a(2a + 1)

= 3a^{2} + a + 2 – 6a^{2} – 34

= -3a^{2} – 2a + 2

(iv) x(x + 4) + 3x(2x^{2} – 1) + 4x^{2} + 4

= x^{2} + 4x + 6x^{3} – 3x + 4x^{2} + 4

= 6x^{3} + 5x^{2} + x + 4

(v) a(b – c) – b(c – a) – c(a – b)

= ab – ac – bc + ab – ca + bc

= 2ab – 2ac

(vi) a(b – c) + b(c – a) + c(a – b)

= ab – ac + bc – ab + ac – bc

= 0

(vii) 4ab(a – b) – 6a^{2}(b – b^{2}) – 3b^{2}(2a^{2} – a) + 2ab(b – a)

= 4a^{2}b – 4ab^{2} – 6a^{2}b + 6a^{2}b^{2} – 6a^{2}b^{2} + 3ab^{2} + 2ab^{2} – 2a^{2}b

= 3ab^{2}

(viii) x^{2}(x^{2} + 1) – x^{3}(x + 1) – x(x^{3} – x)

= x^{4} + x^{2} – x^{4} – x^{3} – x^{4} + x^{2}

= 2x^{2} – 2x^{3}

(ix) 2a^{2} + 3a (1 – 2a^{3}) + a(a + 1)

= 2a^{2} + 3a – 6a^{4} + a^{2} + a

= -6a^{4} + 3a^{2} + 4a

(x) a^{2}(2a – 1) + 3a + a^{3} – 8

= 2a^{3} – a^{2} + 3a + a^{3} – 8

= 3a^{3} – a^{2} + 3a – 8

(xii) a^{2}b(a – b^{2}) + ab^{2}(4ab – 2a^{2}) – a^{3}b(1 – 2b)

= a^{3}b – a^{2}b^{3} + 4a^{2}b^{3} – 2a^{3}b^{2} – a^{3}b + 2a^{3}b^{2}

= -a^{2}b^{3} + 4a^{2}b^{3}

= 3a^{2}b^{3}

(xiii) a^{2}b(a^{3} – a + 1) – ab(a^{4} – 2a^{2} + 2a) – b(a^{3} – a^{2} – 1)

= a^{5}b – a^{3}b + a^{2}b – a^{5}b + 2a^{3}b – 2a^{2}b – ba^{3} + a^{2}b + b

= b

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Volume of a rectangular box with l = b = h = 2x is …………. .

NCERT - Mathematics ExemplarMultiply the monomial by the binomial and find the value of each for x=-1, y=0.25 and z=0.005:

(i) 15y^{2} (2-3x)

(ii) -3x (y^{2}+z^{2})

(iii) z^{2} (x-y)

(iv) xz(x+y^{2})

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