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Identities — MATHS STD 8 — Question

ICSE BoardEnglish MediumSTD 8MATHSIdentities3 Marks
Question
If $3x – 4y = 5$ and $xy = 3$, find : $27x^3 – 64y^3.$
✓

Answer

$27x^3 – 64x^3 = (3x)^3 – (4y)^3= (3x − 4y)^3 (3x − 4y)^3 + 3 (3x) (4y) (3x − 4y) ......[ \because a^3 − b^3= (a − b)^3 + 3ab(a − b)]$
$= (5)^3 + 36(xy) (3x − 4y)$
$= 125 + 36 (3) (5)$
$= 125 + 540$
$= 665$

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