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UPTET Paper 1 Model Test 2

Section: Mathematics

Question : 109 of 150

Marks: +1, -0

\frac{(59\times 59\times 59)+(54\times 54\times 54)+(57\times 57\times 57)-3(59)(54)(57)}{(59+54+57)}

Solution:

Concept:

Use the algebraic identity

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

Explanation:

Let

a=59

b=54

c=57

. The given expression becomes

\frac{a^{3}+b^{3}+c^{3}-3abc}{a+b+c}

Applying the identity, the numerator equals

(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)

. Cancelling

(a+b+c)

gives

a^{2}+b^{2}+c^{2}-ab-bc-ca

Compute squares:

59^{2}=3481

54^{2}=2916

57^{2}=3249

. Sum

=9646

Compute products:

59\times54=3186

54\times57=3078

57\times59=3363

. Sum

=9627

Subtract:

9646-9627=19

(Alternative: Using

a^{3}+b^{3}+c^{3}-3abc=\frac12(a+b+c)[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]

gives the same result after dividing by the denominator.)

Answer:

19

, option D.

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