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a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca): given two parts, find the third

This type of question, asked 10 times: 19 Jul 2023 S2, 19 Jul 2023 S4, 20 Jul 2023 S1, 20 Jul 2023 S3, 21 Jul 2023 S1, 25 Jul 2023 S4, 26 Jul 2023 S2, 26 Jul 2023 S4, 27 Jul 2023 S2, 10 Sep 2024 S1. All 10 questions below, oldest first.

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  1. Question 1

    19 Jul 2023 S2

    54
    If (a+b+c)=12(a + b + c) = 12, and (a2+b2+c2)=50(a^2 + b^2 + c^2) = 50, find the value of (a3+b3+c3−3abc)(a^3 + b^3 + c^3 - 3abc).

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  2. Question 2

    19 Jul 2023 S4

    71
    If (a+b+c)=20(a + b + c) = 20, and a2+b2+c2=152a^2 + b^2 + c^2 = 152, find the value of a3+b3+c3−3abca^3 + b^3 + c^3 - 3abc.

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  3. Question 3

    20 Jul 2023 S1

    67
    If (a3+b3+c3−3abc)=405(a^3 + b^3 + c^3 - 3abc) = 405, and (a+b+c)=15(a + b + c) = 15, find the value of (a−b)2+(b−c)2+(c−a)2(a - b)^2 + (b - c)^2 + (c - a)^2.

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  4. Question 4

    20 Jul 2023 S3

    65
    If a+b+c=7a + b + c = 7, ab+bc+ca=11ab + bc + ca = 11 and abc=−1abc = -1, then a3+b3+c3a^3 + b^3 + c^3 is equal to:

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  5. Question 5

    21 Jul 2023 S1

    58
    If (a+b+c)=12(a + b + c) = 12, and (ab+bc+ca)=47(ab + bc + ca) = 47, find the value of (a3+b3+c3−3abc)(a^3 + b^3 + c^3 - 3abc).

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  6. Question 6

    25 Jul 2023 S4

    67
    If (a+b−c)=20(a + b - c) = 20, and a2+b2+c2=152a^2 + b^2 + c^2 = 152, find the value of a3+b3−c3+3abca^3 + b^3 - c^3 + 3abc.

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  7. Question 7

    26 Jul 2023 S2

    55
    If (a3+b3+c3−3abc)=405(a^3 + b^3 + c^3 - 3abc) = 405, and (a−b)2+(b−c)2+(c−a)2=54(a - b)^2 + (b - c)^2 + (c - a)^2 = 54, find the value of (a+b+c)(a + b + c).

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  8. Question 8

    26 Jul 2023 S4

    55
    If (a+b+c)=14(a + b + c) = 14, and (a3+b3+c3−3abc)=98(a^3 + b^3 + c^3 - 3abc) = 98, find the value of (ab+bc+ca)(ab + bc + ca).

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  9. Question 9

    27 Jul 2023 S2

    65
    If (a+b+c)=14(a + b + c) = 14, and (a3+b3+c3−3abc)=98(a^3 + b^3 + c^3 - 3abc) = 98, find the value of (a2+b2+c2)(a^2 + b^2 + c^2).

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  10. Question 10

    10 Sep 2024 S1

    72
    If the sum of three numbers is 18 and the sum of their squares is 36, then find the difference between the sum of their cubes and three times of their product.

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