Evaluatei -123-73-53ii 283-153-133

(i) nbsp;( ndash;12)3 nbsp;+ 73 nbsp;+ 53 We #160;knowx3+y3+z3 3xyz=x+y+zx2+y2+z2 xy yz zxx3+y3+z3=x+y+zx2+y2+z2 xy yz zx+3xyzHere, #160;x= 12,y=7,z ...

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Byju's Answer

Standard XII

Mathematics

First Fundamental Theorem of Calculus

Evaluatei -12...

Question

Evaluate
(i) (–12)³+ 7³+ 5³
(ii) (28)³ + (–15)³ + (–13)³

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Solution

(i) (–12)³+ 7³+ 5³
$We know x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x) x^{3} + y^{3} + z^{3} = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x) + 3 x y z Here, x = (- 12), y = 7, z = 5$
${(- 12)}^{3} + 7^{3} + 5^{3} = (- 12 + 7 + 5) [{(- 12)}^{2} + 7^{2} + 5^{2} - 7 (- 12) - 35 + 60] + 3 (- 12) \times 35 = 0 - 1260 = - 1260$

(ii) (28)³ + (–15)³ + (–13)³

$We know x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x) x^{3} + y^{3} + z^{3} = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x) + 3 x y z Here, x = (- 28), y = - 15, z = - 13$
${(28)}^{3} + {(- 15)}^{3} + {(- 13)}^{3} = (28 - 15 - 13) [{(28)}^{2} + {(- 15)}^{2} + {(- 13)}^{2} - 28 (- 15) - (- 15) (- 13) - 28 (- 13)] + 3 \times 28 (- 15) (- 13) = 0 + 16380 = 16380$

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Similar questions

Without actually calculating the cubes, find the value of each of the following:

(i)

(ii)

Q. Question 14
Without actually calculating the cubes, find the value of each of the following:

(i) (- 12)^{3} + (7)^{3} + (5)^{3}

(i i) (28)^{3} + (- 15)^{3} + (- 13)^{3}

Q. Without actually calculates the cubes, find the value of

i)

28^{3} + (- 15)^{3} + (- 13)^{3}

i i)

(- 12)^{3} + 7^{3} + 5^{3}

28^{3} + (- 15)^{3} + (- 13)^{3} =

______.

Q. Evaluate: {(28)³ + (−15)³ + (−13)³}.

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Evaluate (i) (–12)3 + 73 + 53 (ii) (28)3 + (–15)3 + (–13)3

Evaluate
(i) (–12)³+ 7³+ 5³
(ii) (28)³ + (–15)³ + (–13)³